hypothesis tests two sample

Two Sample t-test: Definition, Formula, and Example

A two sample t-test is used to determine whether or not two population means are equal.

This tutorial explains the following:

• The motivation for performing a two sample t-test.
• The formula to perform a two sample t-test.
• The assumptions that should be met to perform a two sample t-test.
• An example of how to perform a two sample t-test.

Two Sample t-test: Motivation

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. Since there are thousands of turtles in each population, it would be too time-consuming and costly to go around and weigh each individual turtle.

Instead, we might take a simple random sample of 15 turtles from each population and use the mean weight in each sample to determine if the mean weight is equal between the two populations:

However, it’s virtually guaranteed that the mean weight between the two samples will be at least a little different. The question is whether or not this difference is statistically significant . Fortunately, a two sample t-test allows us to answer this question.

Two Sample t-test: Formula

A two-sample t-test always uses the following null hypothesis:

• H 0 : μ 1  = μ 2 (the two population means are equal)

The alternative hypothesis can be either two-tailed, left-tailed, or right-tailed:

• H 1 (two-tailed): μ 1  ≠ μ 2 (the two population means are not equal)
• H 1 (left-tailed): μ 1  < μ 2  (population 1 mean is less than population 2 mean)
• H 1 (right-tailed):  μ 1 > μ 2  (population 1 mean is greater than population 2 mean)

We use the following formula to calculate the test statistic t:

Test statistic:  ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 )

where  x 1  and  x 2 are the sample means, n 1 and n 2  are the sample sizes, and where s p is calculated as:

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)

where s 1 2  and s 2 2  are the sample variances.

If the p-value that corresponds to the test statistic t with (n 1 +n 2 -1) degrees of freedom is less than your chosen significance level (common choices are 0.10, 0.05, and 0.01) then you can reject the null hypothesis.

Two Sample t-test: Assumptions

For the results of a two sample t-test to be valid, the following assumptions should be met:

• The observations in one sample should be independent of the observations in the other sample.
• The data should be approximately normally distributed.
• The two samples should have approximately the same variance. If this assumption is not met, you should instead perform Welch’s t-test .
• The data in both samples was obtained using a random sampling method .

Two Sample t-test : Example

Suppose we want to know whether or not the mean weight between two different species of turtles is equal. To test this, will perform a two sample t-test at significance level α = 0.05 using the following steps:

Step 1: Gather the sample data.

Suppose we collect a random sample of turtles from each population with the following information:

• Sample size n 1 = 40
• Sample mean weight  x 1  = 300
• Sample standard deviation s 1 = 18.5
• Sample size n 2 = 38
• Sample mean weight  x 2  = 305
• Sample standard deviation s 2 = 16.7

Step 2: Define the hypotheses.

We will perform the two sample t-test with the following hypotheses:

• H 0 :  μ 1  = μ 2 (the two population means are equal)
• H 1 :  μ 1  ≠ μ 2 (the two population means are not equal)

Step 3: Calculate the test statistic  t .

First, we will calculate the pooled standard deviation s p :

s p = √  (n 1 -1)s 1 2  +  (n 2 -1)s 2 2  /  (n 1 +n 2 -2)  = √  (40-1)18.5 2  +  (38-1)16.7 2  /  (40+38-2)  = 17.647

Next, we will calculate the test statistic  t :

t = ( x 1  –  x 2 )  /  s p (√ 1/n 1  + 1/n 2 ) =  (300-305) / 17.647(√ 1/40 + 1/38 ) =  -1.2508

Step 4: Calculate the p-value of the test statistic  t .

According to the T Score to P Value Calculator , the p-value associated with t = -1.2508 and degrees of freedom = n 1 +n 2 -2 = 40+38-2 = 76 is  0.21484 .

Step 5: Draw a conclusion.

Since this p-value is not less than our significance level α = 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the mean weight of turtles between these two populations is different.

Note:  You can also perform this entire two sample t-test by simply using the Two Sample t-test Calculator .

Additional Resources

The following tutorials explain how to perform a two-sample t-test using different statistical programs:

How to Perform a Two Sample t-test in Excel How to Perform a Two Sample t-test in SPSS How to Perform a Two Sample t-test in Stata How to Perform a Two Sample t-test in R How to Perform a Two Sample t-test in Python How to Perform a Two Sample t-test on a TI-84 Calculator

Featured Posts

Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “Two Sample t-test: Definition, Formula, and Example”

I like the detailed information and simplified in the way I can understand and relate easily. Thank you

It seems a couple of parenthesis is missed at the pooled standard deviation formula. Under square root you have (n1-1)s12 + (n2-1)s22 / (n1+n2-2) but it should be [(n1-1)s12 + (n2-1)s22] / (n1+n2-2) I used square bracket

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Join the Statology Community

Sign up to receive Statology's exclusive study resource: 100 practice problems with step-by-step solutions. Plus, get our latest insights, tutorials, and data analysis tips straight to your inbox!

By subscribing you accept Statology's Privacy Policy.

JMP | Statistical Discovery.™ From SAS.

Statistics Knowledge Portal

A free online introduction to statistics

The Two-Sample t -Test

What is the two-sample t -test.

The two-sample t -test (also known as the independent samples t -test) is a method used to test whether the unknown population means of two groups are equal or not.

Is this the same as an A/B test?

Yes, a two-sample t -test is used to analyze the results from A/B tests.

When can I use the test?

You can use the test when your data values are independent, are randomly sampled from two normal populations and the two independent groups have equal variances.

What if I have more than two groups?

Use a multiple comparison method. Analysis of variance (ANOVA) is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if the variances for my two groups are not equal?

You can still use the two-sample t- test. You use a different estimate of the standard deviation. 

What if my data isn’t nearly normally distributed?

If your sample sizes are very small, you might not be able to test for normality. You might need to rely on your understanding of the data. When you cannot safely assume normality, you can perform a nonparametric test that doesn’t assume normality.

See how to perform a two-sample t -test using statistical software

• Download JMP to follow along using the sample data included with the software.
• To see more JMP tutorials, visit the JMP Learning Library .

Using the two-sample t -test

The sections below discuss what is needed to perform the test, checking our data, how to perform the test and statistical details.

What do we need?

For the two-sample t -test, we need two variables. One variable defines the two groups. The second variable is the measurement of interest.

We also have an idea, or hypothesis, that the means of the underlying populations for the two groups are different. Here are a couple of examples:

• We have students who speak English as their first language and students who do not. All students take a reading test. Our two groups are the native English speakers and the non-native speakers. Our measurements are the test scores. Our idea is that the mean test scores for the underlying populations of native and non-native English speakers are not the same. We want to know if the mean score for the population of native English speakers is different from the people who learned English as a second language.
• We measure the grams of protein in two different brands of energy bars. Our two groups are the two brands. Our measurement is the grams of protein for each energy bar. Our idea is that the mean grams of protein for the underlying populations for the two brands may be different. We want to know if we have evidence that the mean grams of protein for the two brands of energy bars is different or not.

Two-sample t -test assumptions

To conduct a valid test:

• Data values must be independent. Measurements for one observation do not affect measurements for any other observation.
• Data in each group must be obtained via a random sample from the population.
• Data in each group are normally distributed .
• Data values are continuous.
• The variances for the two independent groups are equal.

For very small groups of data, it can be hard to test these requirements. Below, we'll discuss how to check the requirements using software and what to do when a requirement isn’t met.

Two-sample t -test example

One way to measure a person’s fitness is to measure their body fat percentage. Average body fat percentages vary by age, but according to some guidelines, the normal range for men is 15-20% body fat, and the normal range for women is 20-25% body fat.

Our sample data is from a group of men and women who did workouts at a gym three times a week for a year. Then, their trainer measured the body fat. The table below shows the data.

Table 1: Body fat percentage data grouped by gender

You can clearly see some overlap in the body fat measurements for the men and women in our sample, but also some differences. Just by looking at the data, it's hard to draw any solid conclusions about whether the underlying populations of men and women at the gym have the same mean body fat. That is the value of statistical tests – they provide a common, statistically valid way to make decisions, so that everyone makes the same decision on the same set of data values.

Checking the data

Let’s start by answering: Is the two-sample t -test an appropriate method to evaluate the difference in body fat between men and women?

• The data values are independent. The body fat for any one person does not depend on the body fat for another person.
• We assume the people measured represent a simple random sample from the population of members of the gym.
• We assume the data are normally distributed, and we can check this assumption.
• The data values are body fat measurements. The measurements are continuous.
• We assume the variances for men and women are equal, and we can check this assumption.

Before jumping into analysis, we should always take a quick look at the data. The figure below shows histograms and summary statistics for the men and women.

The two histograms are on the same scale. From a quick look, we can see that there are no very unusual points, or outliers . The data look roughly bell-shaped, so our initial idea of a normal distribution seems reasonable.

Examining the summary statistics, we see that the standard deviations are similar. This supports the idea of equal variances. We can also check this using a test for variances.

Based on these observations, the two-sample t -test appears to be an appropriate method to test for a difference in means.

How to perform the two-sample t -test

For each group, we need the average, standard deviation and sample size. These are shown in the table below.

Table 2: Average, standard deviation and sample size statistics grouped by gender

Without doing any testing, we can see that the averages for men and women in our samples are not the same. But how different are they? Are the averages “close enough” for us to conclude that mean body fat is the same for the larger population of men and women at the gym? Or are the averages too different for us to make this conclusion?

We'll further explain the principles underlying the two sample t -test in the statistical details section below, but let's first proceed through the steps from beginning to end. We start by calculating our test statistic. This calculation begins with finding the difference between the two averages:

$ 22.29 - 14.95 = 7.34 $

This difference in our samples estimates the difference between the population means for the two groups.

Next, we calculate the pooled standard deviation. This builds a combined estimate of the overall standard deviation. The estimate adjusts for different group sizes. First, we calculate the pooled variance:

$ s_p^2 = \frac{((n_1 - 1)s_1^2) + ((n_2 - 1)s_2^2)} {n_1 + n_2 - 2} $

$ s_p^2 = \frac{((10 - 1)5.32^2) + ((13 - 1)6.84^2)}{(10 + 13 - 2)} $

$ = \frac{(9\times28.30) + (12\times46.82)}{21} $

$ = \frac{(254.7 + 561.85)}{21} $

$ =\frac{816.55}{21} = 38.88 $

Next, we take the square root of the pooled variance to get the pooled standard deviation. This is:

$ \sqrt{38.88} = 6.24 $

We now have all the pieces for our test statistic. We have the difference of the averages, the pooled standard deviation and the sample sizes.  We calculate our test statistic as follows:

$ t = \frac{\text{difference of group averages}}{\text{standard error of difference}} = \frac{7.34}{(6.24\times \sqrt{(1/10 + 1/13)})} = \frac{7.34}{2.62} = 2.80 $

To evaluate the difference between the means in order to make a decision about our gym programs, we compare the test statistic to a theoretical value from the t- distribution. This activity involves four steps:

• We decide on the risk we are willing to take for declaring a significant difference. For the body fat data, we decide that we are willing to take a 5% risk of saying that the unknown population means for men and women are not equal when they really are. In statistics-speak, the significance level, denoted by α, is set to 0.05. It is a good practice to make this decision before collecting the data and before calculating test statistics.
• We calculate a test statistic. Our test statistic is 2.80.
• We find the theoretical value from the t- distribution based on our null hypothesis which states that the means for men and women are equal. Most statistics books have look-up tables for the t- distribution. You can also find tables online. The most likely situation is that you will use software and will not use printed tables. To find this value, we need the significance level (α = 0.05) and the degrees of freedom . The degrees of freedom ( df ) are based on the sample sizes of the two groups. For the body fat data, this is: $ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $ The t value with α = 0.05 and 21 degrees of freedom is 2.080.
• We compare the value of our statistic (2.80) to the t value. Since 2.80 > 2.080, we reject the null hypothesis that the mean body fat for men and women are equal, and conclude that we have evidence body fat in the population is different between men and women.

Statistical details

Let’s look at the body fat data and the two-sample t -test using statistical terms.

Our null hypothesis is that the underlying population means are the same. The null hypothesis is written as:

$ H_o:  \mathrm{\mu_1} =\mathrm{\mu_2} $

The alternative hypothesis is that the means are not equal. This is written as:

$ H_o:  \mathrm{\mu_1} \neq \mathrm{\mu_2} $

We calculate the average for each group, and then calculate the difference between the two averages. This is written as:

$\overline{x_1} -  \overline{x_2} $

We calculate the pooled standard deviation. This assumes that the underlying population variances are equal. The pooled variance formula is written as:

The formula shows the sample size for the first group as n 1 and the second group as n 2 . The standard deviations for the two groups are s 1 and s 2 . This estimate allows the two groups to have different numbers of observations. The pooled standard deviation is the square root of the variance and is written as s p .

What if your sample sizes for the two groups are the same? In this situation, the pooled estimate of variance is simply the average of the variances for the two groups:

$ s_p^2 = \frac{(s_1^2 + s_2^2)}{2} $

The test statistic is calculated as:

$ t = \frac{(\overline{x_1} -\overline{x_2})}{s_p\sqrt{1/n_1 + 1/n_2}} $

The numerator of the test statistic is the difference between the two group averages. It estimates the difference between the two unknown population means. The denominator is an estimate of the standard error of the difference between the two unknown population means. 

Technical Detail: For a single mean, the standard error is $ s/\sqrt{n} $  . The formula above extends this idea to two groups that use a pooled estimate for s (standard deviation), and that can have different group sizes.

We then compare the test statistic to a t value with our chosen alpha value and the degrees of freedom for our data. Using the body fat data as an example, we set α = 0.05. The degrees of freedom ( df ) are based on the group sizes and are calculated as:

$ df = n_1 + n_2 - 2 = 10 + 13 - 2 = 21 $

The formula shows the sample size for the first group as n 1 and the second group as n 2 .  Statisticians write the t value with α = 0.05 and 21 degrees of freedom as:

$ t_{0.05,21} $

The t value with α = 0.05 and 21 degrees of freedom is 2.080. There are two possible results from our comparison:

• The test statistic is lower than the t value. You fail to reject the hypothesis of equal means. You conclude that the data support the assumption that the men and women have the same average body fat.
• The test statistic is higher than the t value. You reject the hypothesis of equal means. You do not conclude that men and women have the same average body fat.

t -Test with unequal variances

When the variances for the two groups are not equal, we cannot use the pooled estimate of standard deviation. Instead, we take the standard error for each group separately. The test statistic is:

$ t = \frac{ (\overline{x_1} -  \overline{x_2})}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} $

The numerator of the test statistic is the same. It is the difference between the averages of the two groups. The denominator is an estimate of the overall standard error of the difference between means. It is based on the separate standard error for each group.

The degrees of freedom calculation for the t value is more complex with unequal variances than equal variances and is usually left up to statistical software packages. The key point to remember is that if you cannot use the pooled estimate of standard deviation, then you cannot use the simple formula for the degrees of freedom.

Testing for normality

The normality assumption is more important   when the two groups have small sample sizes than for larger sample sizes.

Normal distributions are symmetric, which means they are “even” on both sides of the center. Normal distributions do not have extreme values, or outliers. You can check these two features of a normal distribution with graphs. Earlier, we decided that the body fat data was “close enough” to normal to go ahead with the assumption of normality. The figure below shows a normal quantile plot for men and women, and supports our decision.

You can also perform a formal test for normality using software. The figure above shows results of testing for normality with JMP software. We test each group separately. Both the test for men and the test for women show that we cannot reject the hypothesis of a normal distribution. We can go ahead with the assumption that the body fat data for men and for women are normally distributed.

Testing for unequal variances

Testing for unequal variances is complex. We won’t show the calculations in detail, but will show the results from JMP software. The figure below shows results of a test for unequal variances for the body fat data.

Without diving into details of the different types of tests for unequal variances, we will use the F test. Before testing, we decide to accept a 10% risk of concluding the variances are equal when they are not. This means we have set α = 0.10.

Like most statistical software, JMP shows the p -value for a test. This is the likelihood of finding a more extreme value for the test statistic than the one observed. It’s difficult to calculate by hand. For the figure above, with the F test statistic of 1.654, the p- value is 0.4561. This is larger than our α value: 0.4561 > 0.10. We fail to reject the hypothesis of equal variances. In practical terms, we can go ahead with the two-sample t -test with the assumption of equal variances for the two groups.

Understanding p-values

Using a visual, you can check to see if your test statistic is a more extreme value in the distribution. The figure below shows a t- distribution with 21 degrees of freedom.

Since our test is two-sided and we have set α = .05, the figure shows that the value of 2.080 “cuts off” 2.5% of the data in each of the two tails. Only 5% of the data overall is further out in the tails than 2.080. Because our test statistic of 2.80 is beyond the cut-off point, we reject the null hypothesis of equal means.

Putting it all together with software

The figure below shows results for the two-sample t -test for the body fat data from JMP software.

The results for the two-sample t -test that assumes equal variances are the same as our calculations earlier. The test statistic is 2.79996. The software shows results for a two-sided test and for one-sided tests. The two-sided test is what we want (Prob > |t|). Our null hypothesis is that the mean body fat for men and women is equal. Our alternative hypothesis is that the mean body fat is not equal. The one-sided tests are for one-sided alternative hypotheses – for example, for a null hypothesis that mean body fat for men is less than that for women.

We can reject the hypothesis of equal mean body fat for the two groups and conclude that we have evidence body fat differs in the population between men and women. The software shows a p -value of 0.0107. We decided on a 5% risk of concluding the mean body fat for men and women are different, when they are not. It is important to make this decision before doing the statistical test.

The figure also shows the results for the t- test that does not assume equal variances. This test does not use the pooled estimate of the standard deviation. As was mentioned above, this test also has a complex formula for degrees of freedom. You can see that the degrees of freedom are 20.9888. The software shows a p- value of 0.0086. Again, with our decision of a 5% risk, we can reject the null hypothesis of equal mean body fat for men and women.

Other topics

If you have more than two independent groups, you cannot use the two-sample t- test. You should use a multiple comparison   method. ANOVA, or analysis of variance, is one such method. Other multiple comparison methods include the Tukey-Kramer test of all pairwise differences, analysis of means (ANOM) to compare group means to the overall mean or Dunnett’s test to compare each group mean to a control mean.

What if my data are not from normal distributions?

If your sample size is very small, it might be hard to test for normality. In this situation, you might need to use your understanding of the measurements. For example, for the body fat data, the trainer knows that the underlying distribution of body fat is normally distributed. Even for a very small sample, the trainer would likely go ahead with the t -test and assume normality.

What if you know the underlying measurements are not normally distributed? Or what if your sample size is large and the test for normality is rejected? In this situation, you can use nonparametric analyses. These types of analyses do not depend on an assumption that the data values are from a specific distribution. For the two-sample t ­-test, the Wilcoxon rank sum test is a nonparametric test that could be used.

Have a language expert improve your writing

Run a free plagiarism check in 10 minutes, generate accurate citations for free.

• Knowledge Base

Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

• State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a  or H 1 ).
• Collect data in a way designed to test the hypothesis.
• Perform an appropriate statistical test .
• Decide whether to reject or fail to reject your null hypothesis.
• Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Table of contents

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

• H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

Here's why students love Scribbr's proofreading services

Discover proofreading & editing

For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

• an estimate of the difference in average height between the two groups.
• a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

Prevent plagiarism. Run a free check.

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

• Normal distribution
• Descriptive statistics
• Measures of central tendency
• Correlation coefficient

Methodology

• Cluster sampling
• Stratified sampling
• Types of interviews
• Cohort study
• Thematic analysis

Research bias

• Implicit bias
• Cognitive bias
• Survivorship bias
• Availability heuristic
• Nonresponse bias
• Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

Cite this Scribbr article

If you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator.

Bevans, R. (2023, June 22). Hypothesis Testing | A Step-by-Step Guide with Easy Examples. Scribbr. Retrieved July 6, 2024, from https://www.scribbr.com/statistics/hypothesis-testing/

Is this article helpful?

Rebecca Bevans

Other students also liked, choosing the right statistical test | types & examples, understanding p values | definition and examples, what is your plagiarism score.

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

What is a Hypothesis Test for 2 Samples?

Searching the internet for a definition of hypothesis testing for 2 samples brings back a lot of different results. Most of them are a little different. The definitions you will find online usually are disjointed, covering hypothesis testing for independent means, paired means, and proportions. Instead of giving one uniform definition, we’ll take a look at key components that are common to all of the tests, and then some of the specific components and notation.

The Basic Idea

The appearance of these hypothesis tests (in the real world) will be very similar to the tests that we see with one sample. In fact, the examples of hypothesis tests that were in the previous introduction include tests for one sample as well as two samples. The basic structure of these hypothesis tests are very similar to the ones we saw before. You have a problem, hypothesis, data collection, some computations, results or conclusions. Some of the notation will be slightly different. These examples below are the same ones we presented in the previous introduction, but here we are highlighting the two-sample variations. The examples with bolded terms are the ones that use 2 samples.

Some Examples of Hypothesis Tests

Example 1: agility testing in youth football (soccer)players; evaluating reliability, validity, and correlates of newly developed testing protocols.

Reactive agility (RAG)and change of direction speed (CODS) were analyzed in 13U and 15U youth soccer players. “ Independent samples t-test indicated significant differences between U13 and U15 in S10 (t-test: 3.57, p < 0.001), S20M (t-test: 3.13, p < 0.001), 20Y (t-test: 4.89, p < 0.001), FS_RAG (t-test: 3.96, p < 0.001), and FS_CODS (t-test: 6.42, p < 0.001), with better performance in U15. Starters outperformed non-starters in most capacities among U13, but only in FS_RAG among U15 (t-test: 1.56, p < 0.05).”

Most of this might seem like gibberish for now, but essentially the two groups were analyzed and compared, with significant differences observed between the groups. This is a hypothesis test for 2 means, independent samples.

Source: https://pubmed.ncbi.nlm.nih.gov/31906269/

Example 2: Manual therapy in the treatment of carpal tunnel syndrome in diabetic patients: A randomized clinical trial

Thirty diabetic patients with carpal tunnel syndrome were split up into two groups. One received physiotherapy modality and the other received manual therapy. “ Paired t-test revealed that all of the outcome measures had a significant change in the manual therapy group, whereas only the VAS and SSS changed significantly in the modality group at the end of 4 weeks. Independent t-test showed that the variables of SSS, FSS and MNT in the manual therapy group improved significantly greater than the modality group.”

This is a hypothesis test for matched pairs, sometimes known as 2 means, dependent samples.

Source: https://pubmed.ncbi.nlm.nih.gov/30197774/

Example 3: Omega-3 fatty acids decreased irritability of patients with bipolar disorder in an add-on, open label study

“The initial mean was 63.51 (SD 34.17), indicating that on average, subjects were irritable for about six of the previous ten days. The mean for the last recorded percentage was less than half of the initial score: 30.27 (SD 34.03). The decrease was found to be statistically significant using a paired sample t-test (t = 4.36, 36 df, p < .001).”

Source: https://nutritionj.biomedcentral.com/articles/10.1186/1475-2891-4-6

Example 4: Evaluating the Efficacy of COVID-19 Vaccines

“We reduced all values of vaccine efficacy by 30% to reflect the waning of vaccine efficacy against each endpoint over time. We tested the null hypothesis that the vaccine efficacy is 0% versus the alternative hypothesis that the vaccine efficacy is greater than 0% at the nominal significance level of 2.5%.”

Source: https://www.medrxiv.org/content/10.1101/2020.10.02.20205906v2.full

Example 5: Social Isolation During COVID-19 Pandemic. Perceived Stress and Containment Measures Compliance Among Polish and Italian Residents

“The Polish group had a higher stress level than the Italian group (mean PSS-10 total score 22,14 vs 17,01, respectively; p < 0.01). There was a greater prevalence of chronic diseases among Polish respondents. Italian subjects expressed more concern about their health, as well as about their future employment. Italian subjects did not comply with suggested restrictions as much as Polish subjects and were less eager to restrain from their usual activities (social, physical, and religious), which were more often perceived as “most needed matters” in Italian than in Polish residents.”

Even though the test wording itself does not explicitly state the tests we will study, this is a comparison of means from two different groups, so this is a test for two means, independent samples.

Source: https://www.frontiersin.org/articles/10.3389/fpsyg.2021.673514/full

Example 6: A Comparative Analysis of Student Performance in an Online vs. Face-to-Face Environmental Science Course From 2009 to 2016

“The independent sample t-test showed no significant difference in student performance between online and F2F learners with respect to gender [t(145) = 1.42, p = 0.122].”

Once again, a test of 2 means, independent samples.

Source: https://www.frontiersin.org/articles/10.3389/fcomp.2019.00007/full

But what does it all mean?

That’s what comes next. The examples above span a variety of different types of hypothesis tests. Within this chapter we will take a look at some of the terminology, formulas, and concepts related to Hypothesis Testing for 2 Samples.

Key Terminology and Formulas

Hypothesis: This is a claim or statement about a population, usually focusing on a parameter such as a proportion (%), mean, standard deviation, or variance. We will be focusing primarily on the proportion and the mean.

Hypothesis Test: Also known as a Significance Test or Test of Significance , the hypothesis test is the collection of procedures we use to test a claim about a population.

Null Hypothesis: This is a statement that the population parameter (such as the proportion, mean, standard deviation, or variance) is equal to some value. In simpler terms, the Null Hypothesis is a statement that “nothing is different from what usually happens.” The Null Hypothesis is usually denoted by [latex]H_{0}[/latex], followed by other symbols and notation that describe how the parameter from one population or group is the same as the parameter from another population or group.

Alternative Hypothesis: This is a statement that the population parameter (such as the proportion, mean, standard deviation, or variance) is somehow different the value involved in the Null Hypothesis. For our examples, “somehow different” will involve the use of [latex] [/latex], or [latex]\neq[/latex]. In simpler terms, the Alternative Hypothesis is a statement that “something is different from what usually happens.” The Alternative Hypothesis is usually denoted by [latex]H_{1}[/latex], [latex]H_{A}[/latex], or [latex]H_{a}[/latex], followed by other symbols and notation that describe how the parameter from one population or group is different from the parameter from another population or group.

Significance Level: We previous learned about the significance level as the “left over” stuff from the confidence level. This is still true, but we will now focus more on the significance level as its own value, and we will use the symbol alpha, [latex]\alpha[/latex]. This looks like a lowercase “a,” or a drawing of a little fish. The significance level [latex]\alpha[/latex] is the probability of rejecting the null hypothesis when it is actually true (more on what this means in the next section). The common values are still similar to what we had previously, 1%, 5%, and 10%. We commonly write these as decimals instead, 0.01, 0.05, and 0.10.

Test Statistic:  One of the key components of a hypothesis test is what we call a  test statistic . This is a calculation, sort of like a z-score, that is specific to the type of test being conducted. The idea behind a test statistic, relating it back to science projects, would be like calculations from measurements that were taken. In this chapter we will address the test statistic for 2 proportions, 2 means (independent samples), and matched pairs (2 means from dependent samples). The formulas are listed in the table below:

What the different symbols mean:

Critical Region: The critical region , also known as the rejection region , is the area in the normal (or other) distribution in which we reject the null hypothesis. Think of the critical region  like a target area that you are aiming for. If we are able to get a value in this region, it means we have evidence for the claim.

Critical Value: These are like special z-scores for us; the critical value  (or values, sometimes there are two) separates the critical region from the rest of the distribution. This is the non-target part, or what we are not aiming for. If our value is in this region, we do not have evidence for the claim.

P-Value: This is a special value that we compute. If we assume the null hypothesis is true, the p-value represents the probability that a test statistic is at least as extreme as the one we computed from our sample data; for us the test statistics would be either [latex]z[/latex] or [latex]t[/latex].

Decision Rule for Hypothesis Testing:  There are a few ways we can arrive at our decision with a hypothesis test. We can arrive at our conclusion by using confidence intervals, critical values (also known as traditional method), and using p-values. Relating this to a science project, the decision rule would be what we take into consideration to arrive at our conclusion. When we make our decision, the wording will sound a little strange. We’ll say things like “we have enough evidence to reject the null hypothesis” or “there is insufficient evidence to reject the null hypothesis.”

Decision Rule with Critical Values:  If the test statistic is in the critical region, we have enough evidence to reject the null hypothesis. We can also say we have sufficient evidence to support the claim. If the test statistic is not in the critical region, we fail to reject the null hypothesis. We can also say we do not have sufficient evidence to support the claim.

Decision Rule with P-Values: If the p-value is less than or equal to the significance level, we have enough evidence to reject the null hypothesis. We can also say we have sufficient evidence to support the claim. If the p-value is greater than the significance level, we fail to reject the null hypothesis. We can also say we do not have sufficient evidence to support the claim.

More About Hypotheses

Writing the Null and Alternative Hypothesis can be tricky. Here are a few examples of claims followed by the respective hypotheses:

Basic Statistics Copyright © by Allyn Leon is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

Module 10: Hypothesis Testing With Two Samples

Putting it together: hypothesis testing with two samples, let’s summarize.

• The steps for performing a hypothesis test for two population means with unknown standard deviation is generally the same as the steps for conducting a hypothesis test for one population mean with unknown standard deviation, using a t -distribution.
• Because the population standard deviations are not known, the sample standard deviations are used for calculations.
• When the sum of the sample sizes is more than 30, a normal distribution can be used to approximate the student’s  t -distribution.
• The difference of two proportions is approximately normal if there are at least five successes and five failures in each sample.
• When conducting a hypothesis test for a difference of two proportions, the random samples must be independent and the population must be at least ten times the sample size.
• When calculating the standard error for the difference in sample proportions, the pooled proportion must be used.
• When two measurements (samples) are drawn from the same pair of individuals or objects, the differences from the sample are used to conduct the hypothesis test.
• The distribution that is used to conduct the hypothesis test on the differences is a t -distribution.
• Provided by : Lumen Learning. License : CC BY: Attribution
• Introductory Statistics. Authored by : Barbara Illowsky, Susan Dean. Provided by : OpenStax. Located at : https://openstax.org/books/introductory-statistics/pages/1-introduction . License : CC BY: Attribution . License Terms : Access for free at https://openstax.org/books/introductory-statistics/pages/1-introduction

Privacy Policy

Introduction to Statistics and Data Science

Chapter 15 hypothesis testing: two sample tests, 15.1 two sample t test.

We can also use the t test command to conduct a hypothesis test on data where we have samples from two populations. To introduce this lets consider an example from sports analytics. In particular, let us consider the NBA draft and the value of a lottery pick in the draft. Teams which do make the playoffs are entered into a lottery to determine the order of the top picks in the draft for the following year. These top 14 picks are called lottery picks.

Using historical data we might want to investigate the value of a lottery pick against those players who were selected outside the lottery.

We can now make a boxplot comparing the career scoring averages of the lottery picks between these two pick levels.

From this boxplot we notice that the lottery picks tend to have a higher point per game (PPG) average. However, we certainly see many exceptions to this rule. We can also compute the averages of the PTS column for these two groups:

This table once again demonstrates that the lottery picks tend to average more points. However, we might like to test this trend to see if have sufficient evidence to conclude this trend is real (this could also just be a function of sampling error).

15.1.1 Regression analysis

Our first technique for looking for a difference between our two categories is linear regression with a categorical explanatory variable. We fit a regression model of the form: \[PTS=\beta \delta_{\text{ not lottery}}+\alpha\] Where \(\delta_{\text{ not lottery}}\) is equal to one if the draft pick fell outside the lottery and zero otherwise.

To see if this relationship is real we can form a confidence interval for the coefficients.

From this we can see that Lottery picks to tend to average more point per game over their careers. The magnitude of this effect is somewhere between 3.5 and 4.7 points more for lottery picks.

15.1.2 Two Sample t test approach

For this we can use the two-sample t-test to compare the means of these two distinct populations.

Here the alternative hypothesis is that the lottery players score more points \[H_A: \mu_L > \mu_{NL}\] thus the null hypothesis is \[H_0: \mu_L \leq \mu_{NL}.\] We can now perform the test in R using the same t.test command as before.

Notice that I used the magic tilde ~ to split the PTS column into the lottery/non-lottery pick subdivisions. I could also do this manually and get the same answer:

The very small p-value here indicates that the population mean of the lottery picks is truly greater than the population mean of the non-lottery picks.

The 95% confidence interval also tells us that this difference is rather large (at least 3.85 points).

Conditions for using a two-sample t test:

These are roughly the same as the conditions for using a one sample t test, although we now need to assume that BOTH samples satisfy the conditions.

Must be looking for a difference in the population means (averages)

30 or greater samples in both groups (CLT)

• If you have less than 30 in one sample, you can use the t test must you must then assume that the population is roughly mound shaped.

At this point you would probably like to know why we would ever want to do a two sample t test instead of a linear regression?

My answer is that a two sample t test is more robust against a difference in variance between the two groups. Recall, that one of the assumptions of simple linear regression is that the variance of the residuals does not depend on the explanatory variable(s). By default R does a type of t test which does not assume equal variance between the two groups. This is the one advantage of using the t test command.

15.1.2.1 Paired t test

Lets say we are trying to estimate the effect of a new training regiment on the 40 yard dash times for soccer players. Before implementing the training regime we measure the 40 yard dash times of the 30 players. First lets read this data set into R.

First, we can compare the mean times before and after the training:

Also we could make a side by side boxplot for the soccer players times before and after the training

We could do a simple t test to examine whether mean of the players times after the training regime is implemented decrease (on average). Here we have the alternative hypothesis that \(H_a: \mu_b-\mu_a>0\) and thus the null hypothesis that \(H_0: \mu_b-\mu_a \leq 0\) . Using the two sample t test format in R we have:

Here we cannot reject the null hypothesis that the training had no effect on the players sprinting performance. However, we haven’t used all of the information available to us in this scenario. The t test we have just run doesn’t know that we recorded the before and after for the same players more than once. As far as R knows the before and after times could be entirely different players as if we are comparing the results between one team which received the training and one who didn’t. Therefore, R has to be pretty conservative in its predictions. The differences between the two groups could be due to many reasons other than the training regime implemented. Maybe the second set of players just started off being a little bit faster, etc.

The data we collected is actually more powerful because we know the performance of the same players before and after the test. This greatly reduces the number of variables which need to be accounted for in our statistical test. Luckily, we can easily let R know that our data points are paired .

Setting the paired keyword to true lets R know that the two columns should be paired together during the test. We can see that running the a paired t test gives us a much smaller p value. Moreover, we can now safely conclude that the new training regiment is effective in at least modestly reducing the 40 yard dash times of the soccer players.

This is our first example of the huge subject of experimental design which is the study of methods which can be used to create data sets which have more power to distinguish differences between groups. Where possible it is better to collect data for the same subjects under two conditions as this will allow for more powerful statistical analysis of the data (i.e a paired t test instead of a normal t test).

Whenever the assumptions are met for a paired t test, you will be expected to perform a paired t test in this class.

15.2 Two Sample Proportion Tests

We can also use statistical hypothesis testing to compare the proportion between two samples. For example, we might conduct a survey of 100 smokers and 50 non-smokers to see whether they buy organic foods. If we find that 30/100 smokers buy organic and only 11/50 non-smokers buy organic then can we conclude that more smokers buy organic foods that smokers? \(H_a: p_s > p_n\) and \(H_0: p_s \leq p_n\) .

In this case we don’t have sufficient evidence to conclude that a larger fraction of smokers buy organic foods. It is common when analyzing survey data to want to compare proportions between populations.

The key assumptions when performing a two-sample proportion test are that we have at least 5 successes and 5 failures in BOTH samples.

15.3 Extra Example: Birth Weights and Smoking

For this example we are going to use a data from a study on the risk factors associated with giving birth to a low-weight baby (sometimes defined as less than 2,500 grams). This data set is another one which is build into R . To load this data for analysis type:

You can view all a description of the data by typing ?birthwt once it is loaded. To begin we could look at the raw birth weight of mothers who were smokers versus non-smokers. We can do some EDA on this data using a boxplot:

From the boxplot we can see that the median birth weight of babies whose mothers smoked was smaller. We can test the data for a difference in the means using a t.test command.

Notice we can use the ~ shorthand to split the data into those two groups faster than filtering. Here we get a small p value meaning we have sufficient evidence to reject the null hypothesis that the mean weight of babies of women who smoked is greater than or equal to those of non-smokers.

Within this data set we also have a column low which classifies whether the babies birth weight is considered low using the medical criterion (birth weight less than 2,500 grams):

We can see that smoking gives a higher fraction of low-weight births. However, this could just be due to sampling error so let’s run a proportion test to find out.

Once again we find we have sufficient evidence to reject the null hypothesis that smoking does not increase the risk of a low birth weight.

15.4 Homework

15.4.1 concept questions.

• What the assumptions behind using a two sample proportion test? Hint these will be the same as forming a confidence interval for for the fraction of a population, with two samples where this assumption needs to hold.
• What assumptions are required for a two sample t test with small \(N\leq 30\) sample sizes?
• A paired t test may be used for any two sample experiment (True/False)
• The power of any statistical test will increase with increasing sample sizes. (True/False)
• Where possible it is better to collect data on the same individuals when trying to distinguish a difference in the average response to a condition (True/False)
• The paired t test is a more powerful statistical test than a normal t test (True/ False)

15.4.2 Practice Problems

For each of the scenarios below form the null and alternative hypothesis.

• We have conducted an educational study on two classrooms of 30 students using two different teaching methods. The first method had 50% of students pass a standardized test, and the classroom using the second teaching method had 60% of the students pass.
• A basketball coach is extremely superstitious and believes that when he wears his lucky tie the team has a greater chance of winning the game. He comes to you because he is looking to design an experiment to test this belief. If the team has 40 games in the upcoming season, design an experiment and the (null and alt) hypothesis to test the coaches claims.

For the below question work out the number of errors in the data set.

• Before the Olympics all athletes are required to submit a urine sample to be tested for banned substances. This is done by estimating the concentration of certain compounds in the urine and is prone to some degree of laboratory error. In addition, the concentration of these compounds are known to vary with the individual (genetic, diet, etc). To weigh the evidence present in a drug test the laboratory conducts a statistical test. To ensure they don’t falsely convict athletes of doping they use a significance level of \(\alpha=0.01\) . If they test 3000 athletes, all of whom are clean about how many will be falsely accused of doping? Explain the issue with this procedure.

15.4.3 Advanced Problems

Load the drug_use data set from the fivethirtyeight package. Run a hypothesis test to determine if a larger proportion of 22-23 year olds are using marijuana then 24-25 year olds. Interpret your results statistically and practically.

Import the data set Cavaliers_Home_Away_2016 . Form a hypothesis on whether being home or away for the game had an effect on the proportion of games won by the Cavaliers during the 2016-2017 season, test this hypothesis using a hypothesis test.

Load the data set animal_sleep and compare the average total sleep time (sleep_total column) between carnivores and herbivores (using the vore column) to divide the between the two categories. To begin make a boxplot to compare the total sleep time between these two categories. Do we have sufficient evidence to conclude the average total sleep time differs between these groups?

Load the HR_Employee_Attrition data set. We wish to investigate whether the daily rate (pay) has anything to do with whether a employee has quit (the attrition column is “Yes”). To begin make a boxplot of the DailyRate column split into these Attrition categories. Use the boxplot to help form the null hypothesis for your test and decide on an alternative hypothesis. Conduct a statistical hypothesis test to determine if we have sufficient evidence to conclude that those employees who quit tended to be paid less. Report and interpret the p value for your test.

Load the BirdCaptureData data set. Perform a hypothesis test to determine if the proportion of orange-crowned warblers (SpeciesCode==OCWA) caught at the station is truly less than the proportion of Yellow Warblers (SpeciesCode==YWAR). Report your p value and interpret the results statistically and practically.

(All of Statistics Problem) In 1861, 10 essays appeared in the New Orleans Daily Crescent. They were signed “Quintus Curtius Snodgrass” and one hypothesis is that these essays were written by Mark Twain. One way to look for similarity between writing styles is to compare the proportion of three letter words found in two works. For 8 Mark Twain essays we have:

From 10 Snodgrass essays we have that:

• Perform a two sample t test to examine these two data sets for a difference in the mean values. Report your p value and a 95% confidence interval for the results.
• What are the issues with using a t-test on this data?

Consider the analysis of the kidiq data set again.

• Run a regression with kid_score as the response and mom_hs as the explanatory variable and look at the summary() of your results. Notice the p-value which is reported in the last line of the summary. This “F-test” is a hypothesis test with the null hypothesis that the explanatory variable tells us nothing about the value of the response variable.
• Perform a t test for the a difference in means in the kid_score values based on the mom_hs column. What is your conclusion?
• Repeat the t test again using the command:

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

• You've disabled JavaScript in your web browser.
• You're a power user moving through this website with super-human speed.
• You've disabled cookies in your web browser.
• A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.

User Preferences

Content preview.

Arcu felis bibendum ut tristique et egestas quis:

• Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris
• Duis aute irure dolor in reprehenderit in voluptate
• Excepteur sint occaecat cupidatat non proident

Keyboard Shortcuts

7.2.2 - hypothesis testing, derivation of the test section  .

We are now going to develop the hypothesis test for the difference of two proportions for independent samples. The hypothesis test will follow the same six steps we learned in the previous Lesson although they are not explicitly stated.

We will use the sampling distribution of \(\hat{p}_1-\hat{p}_2\) as we did for the confidence interval. One major difference in the hypothesis test is the null hypothesis and assuming the null hypothesis is true.

For a test for two proportions, we are interested in the difference. If the difference is zero, then they are not different (i.e., they are equal). Therefore, the null hypothesis will always be:

\(H_0\colon p_1-p_2=0\)

Another way to look at it is \(H_0\colon p_1=p_2\). This is worth stopping to think about. Remember, in hypothesis testing, we assume the null hypothesis is true. In this case, it means that \(p_1\) and \(p_2\) are equal. Under this assumption, then \(\hat{p}_1\) and \(\hat{p}_2\) are both estimating the same proportion. Think of this proportion as \(p^*\). Therefore, the sampling distribution of both proportions, \(\hat{p}_1\) and \(\hat{p}_2\), will, under certain conditions, be approximately normal centered around \(p^*\), with standard error \(\sqrt{\dfrac{p^*(1-p^*)}{n_i}}\), for \(i=1, 2\).

We take this into account by finding an estimate for this \(p^*\) using the two sample proportions. We can calculate an estimate of \(p^*\) using the following formula:

\(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\)

This value is the total number in the desired categories \((x_1+x_2)\) from both samples over the total number of sampling units in the combined sample \((n_1+n_2)\).

Putting everything together, if we assume \(p_1=p_2\), then the sampling distribution of \(\hat{p}_1-\hat{p}_2\) will be approximately normal with mean 0 and standard error of \(\sqrt{p^*(1-p^*)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\), under certain conditions.

\(z^*=\dfrac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\)

...will follow a standard normal distribution.

Finally, we can develop our hypothesis test for \(p_1-p_2\).

Null: \(H_0\colon p_1-p_2=0\)

Possible Alternatives:

\(H_a\colon p_1-p_2\ne0\)

\(H_a\colon p_1-p_2>0\)

\(H_a\colon p_1-p_2<0\)

Conditions:

\(n_1\hat{p}_1\), \(n_1(1-\hat{p}_1)\), \(n_2\hat{p}_2\), and \(n_2(1-\hat{p}_2)\) are all greater than five

The test statistic is:

\(z^*=\dfrac{\hat{p}_1-\hat{p}_2-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}}\)

...where \(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}\).

The critical values, rejection regions, p-values, and decisions will all follow the same steps as those from a hypothesis test for a one sample proportion.

Example 7-2: Received $100 by Mistake Section  

Let's continue with the question that was asked previously.

Males and females were asked about what they would do if they received a $100 bill by mail, addressed to their neighbor, but wrongly delivered to them. Would they return it to their neighbor? Of the 69 males sampled, 52 said “yes” and of the 131 females sampled, 120 said “yes.”

Does the data indicate that the proportions that said “yes” are different for males and females at a 5% level of significance? Conduct the test using the p-value approach.

• Using Minitab

Again, let’s define males as sample 1.

The conditions are all satisfied as we have shown previously.

The null and alternative hypotheses are:

\(H_0\colon p_1-p_2=0\) vs \(H_a\colon p_1-p_2\ne 0\)

The test statistic:

\(n_1=69\), \(\hat{p}_1=\frac{52}{69}\)

\(n_2=131\), \(\hat{p}_2=\frac{120}{131}\)

\(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{52+120}{69+131}=\dfrac{172}{200}=0.86\)

\(z^*=\dfrac{\hat{p}_1-\hat{p}_2-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\dfrac{\dfrac{52}{69}-\dfrac{120}{131}}{\sqrt{0.86(1-0.86)\left(\frac{1}{69}+\frac{1}{131}\right)}}=-3.1466\)

The p-value of the test based on the two-sided alternative is:

\(\text{p-value}=2P(Z>|-3.1466|)=2P(Z>3.1466)=2(0.0008)=0.0016\)

Since our p-value of 0.0016 is less than our significance level of 5%, we reject the null hypothesis. There is enough evidence to suggest that proportions of males and females who would return the money are different.

  Minitab: Inference for Two Proportions with Independent Samples

To conduct inference for two proportions with an independent sample in Minitab...

The following window will appear. In the drop-down choose ‘Summarized data’ and entered the number of events and trials for both samples.

You should get the following output for this example:

Test and CI for Two Proportions

Difference = p (1) - p (2)

Estimate for difference: -0.162407

95% CI for difference: (-0.274625, -0.0501900)

Test for difference = 0 (vs  ≠ 0): Z = -3.15 P-Value = 0.002 (Use this!)

Fisher's exact test: P-Value = 0.003 (Ignore the Fisher's exact test. This test uses a different method to calculate a test statistic from the Z-test we have learned in this lesson.)

Ignore the Fisher's p -value! The p -value highlighted above is calculated using the methods we learned in this lesson. The Fisher's test uses a different method than what we explained in this lesson to calculate a test statistic and p -value. This method incorporates a log of the ratio of observed to expected values. It's just a different technique that is more complicated to do by-hand. Minitab automatically includes both results in its output.

Try it! Section  

In 1980, of 750 men 20-34 years old, 130 were found to be overweight. Whereas, in 1990, of 700 men, 20-34 years old, 160 were found to be overweight.

At the 5% significance level, do the data provide sufficient evidence to conclude that, for men 20-34 years old, a higher percentage were overweight in 1990 than 10 years earlier? Conduct the test using the p-value approach.

Let’s define 1990 as sample 1.

\(H_0\colon p_1-p_2=0\) vs \(H_a\colon p_1-p_2>0\)

\(n_1=700\), \(\hat{p}_1=\frac{160}{700}\)

\(n_2=750\), \(\hat{p}_2=\frac{130}{750}\)

\(\hat{p}^*=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{160+130}{700+750}=\dfrac{290}{1450}=0.2\)

The conditions are all satisfied: \(n_1\hat{p}_1\), \(n_1(1-\hat{p}_1)\), \(n_2\hat{p}_2\), and \(n_2(1-\hat{p}_2)\) are all greater than 5.

\(z^*=\dfrac{\hat{p}_1-\hat{p}_2-0}{\sqrt{\hat{p}^*(1-\hat{p}^*)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}=\dfrac{\dfrac{160}{700}-\dfrac{130}{750}}{\sqrt{0.2(1-0.2)\left(\frac{1}{700}+\frac{1}{750}\right)}}=2.6277\)

The p-value of the test based on the right-tailed alternative is:

\(\text{p-value}=P(Z>2.6277)=0.0043\)

Since our p-value of 0.0043 is less than our significance level of 5%, we reject the null hypothesis. There is enough evidence to suggest that the proportion of males overweight in 1990 is greater than the proportion in 1980.

  Using Minitab

To conduct inference for two proportions with independent samples in Minitab...

• Choose Stat > Basic Statistics > 2 proportions
• Choose Options

Select "Difference < hypothesized difference" for 'Alternative Hypothesis.

You should get the following output.

Estimate for difference: -0.0552381

95% upper bound for difference: -0.0206200

Test for difference = 0 (vs < 0): Z = -2.63 P-Value = 0.004

Fisher's exact test: P-Value = 0.005 (Ignore the Fisher's exact test)

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

• You've disabled JavaScript in your web browser.
• You're a power user moving through this website with super-human speed.
• You've disabled cookies in your web browser.
• A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.

Snapsolve any problem by taking a picture. Try it in the Numerade app?

Full-Length SAT Suite Practice Tests

Find full-length practice tests on Bluebook™ as well as downloadable paper (nonadaptive) practice tests to help you prepare for the SAT, PSAT/NMSQT, PSAT 10, and PSAT 8/9.

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. The statement is true.

Yes, that is correct. Hypothesis testing involves collecting sample evidence and using probability theory to analyze the likelihood of the hypothesis being true. The process involves setting up a null hypothesis , which is the statement being tested, and an alternative hypothesis. The sample evidence is then gathered and analyzed using statistical tests to determine the probability of observing the results if the null hypothesis were true. If the probability is sufficiently low, the null hypothesis is rejected in favour of the alternative hypothesis . The use of probability theory in hypothesis testing allows for a more objective and rigorous approach to drawing conclusions from data.

The complete question is:-

Hypothesis testing is a procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement. True/False

To learn more about hypothesis testing , refer:-

https://brainly.com/question/30588452

Related Questions

If p=3 what is the value of r when r =19 -5p

Step-by-step explanation:

Erin's water bottle holds 665 milliliters Dylan is carrying two water bottles each one hold 0.35 liters who is carrying more water how much more

Dylan is carrying more water, his two water bottles is 700 millilitres

Erin water bottle holds 665 millilitres

Dylan is carrying two water bottles, each bottle is 0.35 litres

0.35 + 0.35

0.7 liters converted to millilitres is

= 0.7 × 1000

= 700 millilitres

Erin's water bottle is 665 millilitres

Dylan two water bottles is a total of 700 millilitres

Hence Dylan is carrying much more water Read more on millilitres here https://brainly.com/question/20293905

A researcher says that her analysis shows that the effect of the independent variable on the dependent variable is statistically significant. This means the effect is:

The statement means that the observed relationship between the independent variable and the dependent variable is "very unlikely to occur by chance."

In statistical analysis, the concept of statistical significance helps determine if the results obtained from a study are reliable and not due to random chance. When an effect is deemed statistically significant, it suggests that the relationship between the independent variable (the variable being manipulated or studied) and the dependent variable (the variable being measured or observed) is likely to exist in the population from which the sample was drawn.

By reaching statistical significance, the researcher can confidently conclude that the observed effect is more than just a random occurrence and has practical implications in the real world, thus lending support to the underlying hypothesis or research question.

Learn more about independent variable: https://brainly.com/question/25223322

(1) data are obtained from a random sample of assets in millions of dollars of 30 credit unions and the sample mean turns out to be 11.091. assume that the sample standard deviation is 14.404. the population is assumed to be normally distributed. (a) find the 90% confidence interval for the mean. write the solution with two decimal places: for example: (x.xx, x.xx).

The 90% confidence interval for the mean is (6.63, 15.55). Lower limit = X - Margin of error = 11.091 - 4.459 ≈ 6.63 and Upper limit = X + Margin of error = 11.091 + 4.459 ≈ 15.55

To find the 90% confidence interval for the mean, we can use the formula: CI = X ± Zα/2 * (σ/√n) Where X is the sample mean (11.091), Zα/2 is the z-score corresponding to the desired confidence level (90% confidence interval = 1.645), σ is the sample standard deviation (14.404), and n is the sample size (30). Plugging in the values, we get: CI = 11.091 ± 1.645 * (14.404/√30) CI = (3.20, 18.98) Therefore, the 90% confidence interval for the mean is (3.20, 18.98).

The 90% confidence interval for the mean . Given the sample size (n = 30), the sample mean (X = 11.091), and the sample standard deviation (s = 14.404), we can calculate the confidence interval using the formula: X ± t * (s / √n) First, we need to find the t-value for a 90% confidence interval with 29 degrees of freedom (n - 1). You can use a t-table or a calculator to find this value. For a 90% confidence interval, the t-value is approximately 1.699. Next, we calculate the margin of error : Margin of error = t * (s / √n) = 1.699 * (14.404 / √30) = 4.459 Now, we can find the confidence interval: Lower limit = X - Margin of error = 11.091 - 4.459 ≈ 6.63 Upper limit = X + Margin of error = 11.091 + 4.459 ≈ 15.55 Thus, the 90% confidence interval for the mean is (6.63, 15.55).

To learn more about confidence interval , click here:

brainly.com/question/24131141

please help i need a good explanation!! thanks!! :)

The sides of the quadrilateral are

When the quadrilateral are similar according to the ratio 7 : 11 and the similar sides are:

Since FJ : AD the ratio is 11 : 7, therefore where FJ = 11 then AD = 7 = x

AB = 2x = 2 * 7 = 14

cross multiplying

7 * FG = 14 * 11

GH ⇔ BC = 3 * 7 = 21

7 * GH = 21 * 11

HJ ⇔ CD = 4 * 7 = 28

7 * GH = 28 * 11

Learn more about similar quadrilateral at

https://brainly.com/question/27991573

a sailfish can travel as fast as 68 miles per hour. at that rate, how far would a sailfish travel in 45 minutes

Therefore, a sailfish would travel 51 miles in 45 minutes if it maintained a speed of 68 miles per hour in the equation .

To calculate the distance traveled in a given time, we can use the formula distance = rate x time, where rate refers to the speed of travel and time refers to the duration of travel. In this scenario, we are given a rate of 68 miles per hour and a time of 45 minutes.

To use the formula , we first need to convert the time to hours since the rate is given in miles per hour. We do this by dividing the time by 60, since there are 60 minutes in an hour. In this case, 45 minutes divided by 60 minutes per hour gives us 0.75 hours.

Now, we can plug in the values for rate and time into the formula and solve for distance. Multiplying 68 miles per hour by 0.75 hours gives us a distance of 51 miles.

45 minutes / 60 minutes per hour = 0.75 hours

Then, we can use the formula: distance = rate x time

distance = 68 miles per hour x 0.75 hours

distance = 51 miles

To know more about equation ,

https://brainly.com/question/28243079

find the conditional probability of the indicated event when two fair dice (one red and one green) are rolled. hint [see example 1.] the sum is 4, given that the green one is either 3 or 2.

The conditional probability of the sum being 4, given that the green die shows either a 3 or a 2, is 1/6.

To find the conditional probability of the sum being 4, given that the green die is either 3 or 2, we need to use the formula: P(A|B) = P(A and B) / P(B) where A is the event of getting a sum of 4 and B is the event of getting either a 3 or 2 on the green die. First, let's calculate the probability of getting a 3 or 2 on the green die: P(B) = 1/3 + 1/3 = 2/3 since there are 3 possible outcomes for each die and the green die can either be 3 or 2. Next, we need to calculate the probability of getting a sum of 4 and a green die of either 3 or 2: P(A and B) = 2/36 since there are only 2 ways to get a sum of 4 with a green die of either 3 or 2: (1,3) and (2,2). Now we can plug in the values into the formula: P(A|B) = (2/36) / (2/3) = 1/18 Therefore, the conditional probability of getting a sum of 4, given that the green die is either 3 or 2, is 1/18. To find the conditional probability of the indicated event, we'll use the formula: P(A / B) = P(A / B) / P(B) Here, event A is the sum of the numbers on the two dice being 4, and event B is the green die showing either a 3 or a 2. First, let's find P(B). There are 6 possible outcomes for each die, so there are 6x6=36 total possible outcomes when rolling both dice. There are 2 favorable outcomes for event B: the green die showing a 3 or a 2. Therefore, P(B) = 2/6 = 1/3. Now, let's find P(A / B). This is the probability of both events A and B happening at the same time. For the sum to be 4 and the green die to show a 2, the red die must show a 2. For the sum to be 4 and the green die to show a 3, the red die must show a 1. There are 2 favorable outcomes for P(A /B) out of the 36 possible outcomes. Therefore, P(A ∩ B) = 2/36 = 1/18. Finally, we can find the conditional probability P(A | B) using the formula: P(A / B) = P(A / B) / P(B) = (1/18) / (1/3) = (1/18) * (3/1) = 3/18 = 1/6. So, the conditional probability of the sum being 4, given that the green die shows either a 3 or a 2, is 1/6.

To learn more about probability , click here:

brainly.com/question/30034780

We calculated that the value of the test statistic is z = 3.53. Next we will calculate the P-value. Since the alternative hypothesis is of the form p > Po, the P-value is the area under the z curve to the right of the calculated value of the test statistic. Using SALT, we find that the cumulative probability associated with a z test statistic of 3.53, rounded to four decimal places, is 1.6449. We can use the cumulative probability to find the P-value. (Round your answers to four decimal places.)P-value = (z curve area to the right of 3.53) = P(z < 3.53) = 1 1.6449 xP Value is =

Cumulative probabilities are associated with the area to the left of the test statistic, you will subtract the cumulative probability from 1 to find the area to the right.P-value = 1 - P(z < 3.53)

Once you have the correct cumulative probability, plug it into the equation and calculate the P-value. After calculating the test statistic z = 3.53, you need to find the P-value . Since the alternative hypothesis is of the form p > Po, the P-value is the area under the z curve to the right of the calculated value of the test statistic. Using SALT, you found the cumulative probability associated with a z test statistic of 3.53, rounded to four decimal places, is 1.6449. However, the value 1.6449 seems to be incorrect as the cumulative probability should be less than 1. To find the P-value, you need to calculate the area under the z curve to the right of z = 3.53.

Know more about probability here:

https://brainly.com/question/30034780

At a track meet, a relay race with a total distance of 1\2 mile is run by a team of 3 runners. Each runner completed the same distance in the race. Which of these models could be used to find the distance, in miles, run by each of the 3 runners? Choose the THREE correct answers.

First models could be used to find the distance , in miles, run by each of the 3 runners.

A relay race with a total distance of 1/2 mile is run by a team of 3 runners.

So, Distance Covered by each runner is

= (1/3) (1/2)  

So, (1/2) x (1/3) = 1/6

(1/2) ÷ 3 = 1/6

So, the correct model   1/2 mile is divided in 3 equal parts are the correct options.

Learn more about Fraction here:

brainly.in/question/5451929

How are databases of variants used to help find disease gene candidates? Variants present in individuals that do not have the disease or are common in the general population are unlikely to cause a rare genetic disease. They can indicate the probability of different variants appearing in a population. Variants from individuals that do not have the disease are not useful for this purpose. They can indicate rare variants that will help with the genetic identification of individuals

Databases of variants are crucial in helping to identify disease gene candidates. Here's how they are used:

1. Collect and store genetic variants: Databases collect and store genetic variants from various individuals, including those with specific diseases and those without.

2. Compare variants between groups: By comparing the variants in affected individuals to those in unaffected individuals, researchers can identify variants that are more common in the disease group. This helps to narrow down the list of potential disease-causing genes.

3. Calculate probability : Databases can be used to calculate the probability of certain variants appearing in the general population. If a variant is rare in the general population but more common in individuals with a specific disease, it may be more likely to be a disease-causing variant.

4. Filter out common variants : As you mentioned, variants that are common in the general population or present in individuals without the disease are less likely to cause a rare genetic disease. By filtering out these common variants, researchers can focus on rare variants that may have a stronger association with the disease. 5. Identify disease gene candidates: Through this process of comparing and filtering genetic variants, researchers can identify potential disease gene candidates that warrant further investigation.

In summary, databases of variants help researchers identify disease gene candidates by storing genetic information, enabling comparison between affected and unaffected individuals, calculating variant probabilities, filtering out common variants, and ultimately pinpointing rare variants that may be associated with specific genetic diseases.

To know more about probability click here

brainly.com/question/15124899

Help with what u can please!

The building is 51 feet tall

The value of x is 1.5

The length of the building is solved by comparing the building's height by John's height as this forms similar triangles

The comparison is done using the proportions as below

John's shadow / John's height = building's shadow / building's height

6 / 4.5 = 68 / building's height

building's height x 6 = 4.5 * 68

building's height = 4.5 x 68 / 6

building's height = 51 feet

Solving for x

Using proportions

15 / 10 = 23x / 20x - 8

(20x - 8) * 15 = 22x * 10

300x - 120 = 220x

300x - 220x = 120

Learn more about similar triangles at

https://brainly.com/question/14285697

In a weighted grading system, students are graded on quizzes, tests, and a project, each with a different weight. Matrix W represents the weights for each kind of work, and matrix G represents the grades for two students, Felipe and Helena. Q T P W = [0.40 0.50 0.10] Felipe Helena G= Q {80 70} T {60 80} p { 90 60} Final grades are represented in a matrix F. If F = WG, what is F? A. [7174] B. [7174] C. [7471] D. [7471]

For Felipe and Helena's final grades, the solution is option C, [74 71].

Using the given values for Q, T, and P weights and Felipe and Helena's grades, calculate their final grades as follows:

Felipe's final grade:

0.40 x 80 + 0.50 x 60 + 0.10 x 90 = 32 + 30 + 9 = 71

Helena's final grade:

0.40 x 70 + 0.50 x 80 + 0.10 x 60 = 28 + 40 + 6 = 74

To represent the final grades for Felipe and Helena in a matrix F , given formula F = WG, where W = matrix of weights and G = matrix of grades:

[0.40 0.50 0.10]   [80 70]

F = WG = [0.40 0.50 0.10] x [60 80]

[0.40 0.50 0.10] [90 60]

Performing matrix multiplication :

[32 + 30 + 9  28 + 40 + 6]

F = WG = [32 + 40 + 6 28 + 40 + 3]

[36 + 25 + 6 36 + 20 + 3]

Simplifying:

F = WG = [78 71]

Therefore, [74 71] for Felipe and Helena's final grades, respectively.

Find out more on matrix here: https://brainly.com/question/2456804

At a particular restaurant, 52% of all customers order an appetizer and 32% of all customers order dessert. If 27% of all customers order both an appetizer and dessert, what is the probability a randomly selected customer orders an appetizer or dessert or both? Write your answer as a decimal (not as a percentage).

The probability of randomly selected c ustomers ordering both an appetizer and dessert is 57%.

The probability is defined as the possibility of an event being equal to the ratio of the number of favorable outcomes and the total number of outcomes.

Given data as:

P(E₁) = 52%

P(E₂) = 32%

P(E₁ or E₂) = 27%.

Using the formula:

⇒ P(E₁ or E₂) = P(E₁) + P(E₂) - P(E₁ & E₂)

Substitute the values and solve for P(E₁ & E₂)

⇒ P(E₁ & E₂) = 52% + 32% - 27%

⇒ P(E₁ & E₂) = 57%

Therefore, the probability of randomly selected customers ordering both an appetizer and dessert is 57%.

Learn more about the probability here:

brainly.com/question/11234923

Which equation represents the parabola with focus ( 1 , 4 ) and latus rectum of 4? y = − 1 4 ( x − 1 ) 2 + 3 y = 4 ( x − 1 ) 2 + 3 y = ( x − 1 ) 2 + 4 y = 1 4 ( x − 1 ) 2 + 3

The equation of parabola with with focus (1,4)and latus rectum of 4 is given as y=(x−1)2+4, hence option C is correct.

We know that the focus of the parabola is at (1, 4), and the latus rectum is 4 units long. The latus rectum is the line segment that passes through the focus and is perpendicular to the axis of symmetry. Its length is equal to 4 times the distance from the focus to the directrix . Since the latus rectum is 4 units long, the distance from the focus to the directrix is 1 unit.

Therefore, the directrix is the horizontal line y = 3, which is 1 unit below the focus. The axis of symmetry is the vertical line x = 1. The standard form of the equation of a parabola with vertex at the origin, axis of symmetry along the x-axis, and focus at (p, 0) is:

In this case, the vertex is (1, 0) and the focus is (1, 4). Therefore, we can write,

x - 1 = 4p(y - 0)

x - 1 = 4py

We know that the distance from the focus to the vertex is p, which is 4 units in this case. Therefore, the equation of the parabola is,

(x - 1)² = 16y

Simplifying this equation , we get,

y = (1/16)(x - 1)²

Therefore, the answer is C. y = (x - 1)²/16.

To know more about equation of parabola , visit,

https://brainly.com/question/4061870

Complete question - Which equation represents the parabola with focus (1,4) and latus rectum of 4?

A. y=−14(x−1)²+3

B. y=4(x−1)²+3

C. y=(x−1)²+4

D. y=14(x−1)²+3

let a represent the average value of the function f(x) on the interval [0.6]. is there a value of c for which the average value of f(x) on the interval [0. c] is greater than a? explain why or why

The average value of a function f(x) on an interval [0, 6] is represented by 'a'. To determine if there is a value 'c' for which the average value of f(x) on the interval [0, c] is greater than 'a', we need to consider the properties of the function and the Mean Value Theorem. The Mean Value Theorem states that if a function f(x) is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point 'c' in the interval (a, b) such that the average rate of change equals the instantaneous rate of change or f'(c) = (f(b) - f(a)) / (b - a). Without more information about the function f(x), we cannot definitively say whether there is a value 'c' for which the average value of f(x) on the interval [0, c] is greater than the average value on the interval [0, 6]. However, if the function meets the conditions of the Mean Value Theorem, it is possible that such a value 'c' exists.

Learn more about average here:

https://brainly.com/question/24057012

Check all statements that are true about the Pythagorean Theorem Only used on Right Triangles The hypotenuse is the shortest side of the triangle The legs are sides a and b of the triangles The last step is to take the square root of both sides Exponents are NOT a part of the pythagorean theorem

The true statements about the Pythagorean theorem are :

a) Only used on Right Triangles

b) The legs are sides a and b of the triangles

c) The last step is to take the square root of both sides

Given data ,

The Pythagorean Theorem is a mathematical theorem that applies only to right triangles , and it states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the two other sides, called the legs.

a² + b² = c²

where a and b are the lengths of the legs, and c is the length of the hypotenuse

To learn more about triangles click :

https://brainly.com/question/16739377

Use the discriminant to determine whether each quadratic equation has two real solutions, a double root, or no real solution, without solving the equation. 2a2 + 3a + 1 = 0 2a(a – 3) – 3 = 3 3a2 – 2a(a – 2)2 – 4 = 0

The discriminant of a quadratic equation is the expression b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c. To determine whether each quadratic equation has two real solutions, a double root, or no real solution, we need to look at the value of the discriminant. For the equation 2a^2 + 3a + 1 = 0, the discriminant is b^2 - 4ac = 3^2 - 4(2)(1) = 1. Since the discriminant is positive and not equal to zero, there are two real solutions. For equation 2a(a – 3) – 3 = 3, we need to rearrange it into standard form first, which gives us 2a^2 - 6a - 6 = 0. The discriminant is b^2 - 4ac = (-6)^2 - 4(2)(-6) = 60. Since the discriminant is positive, there are two real solutions. For the equation 3a^2 – 2a(a – 2)^2 – 4 = 0, we need to expand the squared term first, which gives us 3a^2 - 2a(a^2 - 4a + 4) - 4 = 0. Simplifying this equation gives us 3a^2 - 2a^3 + 8a - 4 = 0. The discriminant is b^2 - 4ac = 8^2 - 4(3)(-2) = 100. Since the discriminant is positive, there are two real solutions. Therefore, all three quadratic equations have two real solutions.

Learn more about quadratic here:

https://brainly.com/question/22364785

How can we easily recognize when a system of linear equations is inconsistent or not?

To recognize linear equations write them in Ax = b form, Gaussian elimination, check in raw-echelon for all coefficients in zero, and if it exists its a contradiction and inconsistent .

To easily recognize when a system of linear equations is inconsistent, you can follow these steps: 1. Write the system of linear equations in the form Ax = b, where A is the matrix of coefficients, x is the vector of variables, and b is the constant vector. 2. Perform Gaussian elimination or row reduction on the augmented matrix [A | b] to obtain the row-echelon form. 3. Check for any row in the row-echelon form where all the coefficients of the variables are zero, but the constant term is nonzero (i.e., 0x + 0y + ... + 0z = k, where k ≠ 0). 4. If such a row exists, then the system of linear equations is inconsistent because it represents a contradiction (e.g., 0 = k, where k ≠ 0). If no such row is found, then the system is either consistent and has a unique solution or consistent and has infinitely many solutions, depending on the number of free variables.

Learn more about linear equations here:

https://brainly.com/question/27878329

Consider a circle whose equation is x2 + y2 – 2x – 8 = 0. Which statements are true? Select three options. The radius of the circle is 3 units. The center of the circle lies on the x-axis. The center of the circle lies on the y-axis. The standard form of the equation is (x – 1)² + y² = 3. The radius of this circle is the same as the radius of the circle whose equation is x² + y² = 9

From the equation of the circle , the radius of the circle is 3 and the center lies at (1, 0)

A circle is a closed curve that is drawn from the fixed point called the center, in which all the points on the curve are having the same distance from the center point of the center. The equation of a circle with (h, k) center and r radius is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation. Thus, if we know the coordinates of the center of the circle and its radius as well, we can easily find its equation.

In this problem, we have an equation x^2 + y^2 - 2x - 8 == 0

The radius of this circle is 3 and the center is at (1, 0).

The correct option are A and C

Learn more on equation of circle here;

https://brainly.com/question/24810873

You owe $1,500 on a credit card with a 14.5% APR. The minimum payment is $80. How much goes toward the principal if you make the minimum payment at the end of the first month?

If you make the minimal payment of $80 on the end of the first month, $61.85 will cross towards the principal and $18.15 will go towards interest charges.

To calculate how much of the minimal charge goes closer to the principal, we first need to calculate the amount of interest that accrues at the outstanding balance for the first month.

Interest charged for the month = monthly interest fee x outstanding stability

Now we are able to calculate how a good deal of the minimal fee goes closer to the principal :

Consequently, if you make the minimal payment of $80 on the end of the first month, $61.85 will cross towards the principal and $18.15 will go towards interest charges.

Learn more about Simple Interest:-

https://brainly.com/question/29716197

Can someone help me asap? It’s due today!! I will give brainliest if it’s correct

Find the missing side of the triangle. 10 cm and 6 cm. What is X??????? I missed my RSM class and now this?!

The missing side of triangle is 11.66 cm, under the condition that the given triangle is a right angled triangle and the other sides of the triangle are 10cm, 6cm respectively.

In order to evaluate the other side of the triangle, we have to rely on the principles of Pythagorean Theorem which states that in a right triangle, the sum of the square sides are equal to the square of the hypotenuse side.

Then, let us consider that x be the length of the missing side then

x² = 10² + 6²

x² = 100 + 36

x ≈ 11.66 cm

Then, the missing side of this triangle is approximately 11.66 cm.

To learn more about right angle triangle

https://brainly.com/question/29869536

The complete question is

Find the missing siside of the following right triangle in the figure

At the beginning of the year, a company estimates total overhead costs of $1,309,620. The company applies overhead using machine hours and estimates that it will use 2,990 machine hours during the year. What amount of overhead should be applied to a job that uses 30 machine hours that year?

If at the beginning of the year, a company estimates total overhead costs of $1,309,620. The amount of overhead should be applied to a job that uses 30 machine hours that year is: 13,140.

Overhead rate per machine hour = Total estimated overhead costs / Estimated number of machine hour

Overhead rate per machine hour = $1,309,620 / 2,990

Overhead rate per machine hour  = $438 per machine hour

Overhead applied to job = Overhead rate per machine hour x Number of machine hours used

Overhead applied to job = $438 x 30

Overhead applied to job = $13,140

Therefore, the amount of overhead is $13,140.

Learn more about overhead here:https://brainly.com/question/13037939

Suppose that we want to estimate what proportions of all drivers exceed the legal speed limit on a certain stretch of road between Los Angeles and Bakersfield. Use the formula of the earlier exercise to determine how large a sample we will need to be at least 99 % confident that the resulting estimate, the sample proportion, is off by less than 0.04.

We need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

To estimate the proportion of all drivers exceeding the legal speed limit on a certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level, we need to use the following formula: [tex]n = (Z^2 * p * (1 - p)) / E^2[/tex] where n is the sample size , Z is the Z-score for the desired confidence level (2.576 for 99% confidence level), p is the estimated proportion of drivers exceeding the speed limit (we don't have an estimate, so we'll use 0.5 for maximum variability), and E is the margin of error we want (0.04). Plugging in the values, we get: [tex]n = (2.576^2 * 0.5 * (1 - 0.5)) / 0.04^2[/tex] n = 664.92 Therefore, we need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

learn more about   confidence level,

https://brainly.com/question/22851322

The student government association at a university wants to estimate the percentage of the student body that supports a change being considered in the academic calendar of the university for the next academic year. How many students should be surveyed if a 99% confidence interval is desired and the margin of error is to be only 8%? The student government should survey Student (Round up to the nearest integer.)

The student government should survey 640 students to estimate the percentage of the student body that supports the proposed change in the academic calendar with a 99% confidence level and a margin of error of 8%.

To determine the sample size needed for the survey, we can use the following formula: n = (z^2 * p * q) / E^2 Where: n = sample size z = z-score for the desired confidence level (99%) p = estimated proportion of the population supporting the change (unknown, so we'll use 0.5 for a conservative estimate ) q = 1 - p E = margin of error (0.08) Plugging in the values, we get: n = (2.576^2 * 0.5 * 0.5) / 0.08^2 n = 639.53 Since we can't have a fractional number of students, we should round up to the nearest integer: n = 640 Learn more about desired confidence level here:

https://brainly.com/question/30078824

What is the ratio of raccoons to total animals?

We can see here that the ratio of   raccoons to total animals is: B. 4:6

We can define ratio in mathematics as a division-based comparison of two numbers or quantities. It conveys how big or much one quantity is in proportion to another.

A fraction is a common way to express ratios , with the first number denoting how much of the first quantity there is, and the second number denoting how much of the second quantity there is.

Learn more about ratio on https://brainly.com/question/12024093

Complete the following tasks on the plane. Using a blue pencil, shade the region that contains points that are more than 2 1/2 units and less than 3 1/4 units from the y-axis.

The points that are more than the fraction 2 1/2 units and less than 3 1/4 units contains the points from 2.5 to 3.25.

Given measurements are 2 1/2 and 3 1/4.

We have to shade the region in between these two points .

Both are written in mixed fractional form.

This can be written in decimal as,

2 1/2 = 2 + 1/2 = 2 +6 0.5 = 2.5

3 1/4 = 3 + 1/4 = 3 + 0.25 = 3.25

Hence the shaded region will contain the points from 2.5 to 3.25.

2.5 is at the exact middle of 2 and 3.

3.25 is at 1/4th distance of 3 and 4. That is if we divide the distance between 3 and 4 to 4 equal spaces, the first space is 3 1/4.

Learn more about Fractions here :

https://brainly.com/question/15993912

If f(x) = 2x - 9, which of the following are correct? Select all that apply. f(-3) = 15 f(-1) = -11 f(0) = -9 f(2) = 5 f(3) = -3

f(0)=-9 and f(3)=-3

f(-3)=15 - so to solve this we have to replace x with -3, which would equal:

2(-3)-9=-15.  -15 doesn't equal 15, so this is incorrect.

f(-1)=-11 Substitute x for -1: 2(-1)-9=-12.  -12 doesn't equal -11, so this is also incorrect.

f(0)=-9 Substitute x for 0.  2(0)-9= -9.  -9 does equal -9, so this is correct.

f(2)=5 Substitute x for 2.  2(2)-9=-5. This isn't correct.

f(3)=-3 Substitute x for 3.  2(3)-9=-3.  This is also correct because -3 does equal -3.

Hope this helps!  :)

rearrange the following steps in the correct order to find the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails. Rank the options below. The probability is 1/16 1/16 Of these, only one will result in four heads appearing, namely THHHH. There are 16 equally likely outcomes of flipping a fair coin five times in which the first flip comes up tails.

The correct order of steps is 1, 2, 3, 4. And the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails , is 1/16.

To find the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails, we need to follow these steps in order: 1. Identify the total number of possible outcomes when a fair coin is flipped five times, which is 2^5 = 32. 2. Determine the number of outcomes in which the first flip is tails, which is also 16. 3. Out of the 16 outcomes where the first flip is tails , identify the number of outcomes in which exactly four heads appear. There is only one such outcome: THHHH. 4. Calculate the conditional probability by dividing the number of favourable outcomes (i.e. THHHH) by the number of total outcomes given the condition (i.e. the first flip is tails), which is 1/16. Therefore, the correct order of steps is 1, 2, 3, 4. And the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails, is 1/16.

To learn more about conditional probability , refer:-

https://brainly.com/question/30144287

a manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 408 gram setting. it is believed that the machine is underfilling or overfilling the bags. a 13 bag sample had a mean of 399 grams with a variance of 121 . assume the population is normally distributed. is there sufficient evidence at the 0.02 level that the bags are underfilled or overfilled?

After the test statistic we find that the critical t-value is 2.681. Since the absolute value of the test statistic (-2.78) is greater than the critical t-value (2.681), we can reject the null hypothesis.

The test statistic can be calculated as follows: t = (399 - 408) / (sqrt(121/13)) = -2.78  Since the absolute value of the test statistic (-2.78) is greater than the critical t-value (2.681), we can reject the null hypothesis.

To answer your question, we will conduct a hypothesis test to determine if there is sufficient evidence at the 0.02 significance level that the bags are underfilled or overfilled.

Step 1: State the hypotheses H0 (null hypothesis): The population means (μ) is 408 grams. H1 (alternative hypothesis): The population means (μ) is not equal to 408 grams. Step 2: Determine the test statistic Since we know the population variance, we will use a z-test. The formula for the z-test statistic is: z = (sample mean - population mean) / (population standard deviation/sqrt (sample size)) z = (399 - 408) / (sqrt(121) / sqrt(13)) Step 3: Calculate the z-value z = (-9) / (11 / sqrt(13)) z ≈ -2.58 Step 4: Determine the critical value For a two-tailed test at the 0.02 significance level , we need to find the critical value. Using a z-table, we find the critical values are approximately -2.33 and +2.33. Step 5: Compare the z-value to the critical values Our calculated z-value (-2.58) is less than the lower critical value (-2.33). Step 6: Draw a conclusion Since our z-value falls in the rejection region, we reject the null hypothesis. There is sufficient evidence at the 0.02 significance level to conclude that the chocolate chip bag-filling machine is either underfilling or overfilling the bags when set to 408 grams, assuming the population is normally distributed.

Learn more about Test Statistic :

brainly.com/question/14128303