hypothesis testing matched sample

10.4 Matched or Paired Samples

When using a hypothesis test for matched or paired samples, the following characteristics should be present:

• Simple random sampling is used.
• Sample sizes are often small.
• Two measurements (samples) are drawn from the same pair of individuals or objects.
• Differences are calculated from the matched or paired samples.
• The differences form the sample that is used for the hypothesis test.
• Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences.

Example 10.11

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.14 . A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.")

The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6}

The sample mean and sample standard deviation of the differences are: x – d = –3.13 x – d = –3.13 and s d = 2.91 s d = 2.91 Verify these values.

Let μ d μ d be the population mean for the differences. We use the subscript d d to denote "differences."

Random variable: X ¯ d X ¯ d = the mean difference of the sensory measurements

H 0 : μ d ≥ 0

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ d is the population mean of the differences.)

H a : μ d < 0

The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t 7 . (Notice that the test is for a single population mean.)

Calculate the test statistic and p -value using the Student's t -distribution:

test statistic, t = –3.036, p -value = 0.0095

X ¯ d X ¯ d is the random variable for the differences.

The sample mean and sample standard deviation of the differences are:

x ¯ d x ¯ d = –3.13

s ¯ d s ¯ d = 2.91

Compare α and the p -value: α = 0.05 and p -value = 0.0095. α > p -value.

Make a decision: Since α > p -value, reject H 0 . This means that μ d < 0 and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.

For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time ( after - before ) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1 st list name - 2 nd list name. The calculator will do the subtraction, and you will have the differences in the third list.

Using the TI-83, 83+, 84, 84+ Calculator

Use your list of differences as the data. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 0 for μ 0 μ 0 , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ : and arrow over to < μ 0 μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate ). Press ENTER .

Try It 10.11

A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level.

Example 10.12

A college softball coach was interested in whether the college's strength development class increased their players' maximum lift (in pounds). Four players participated in the study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:

The coach wants to know if the strength development class makes the players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation.

x ¯ d = 21.3 , s d = 46.7 x ¯ d = 21.3 , s d = 46.7

The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative.

Using the difference data, this becomes a test of a single __________ (fill in the blank).

Define the random variable: X ¯ d X ¯ d mean difference in the maximum lift per player.

The distribution for the hypothesis test is t 3 .

H 0 : μ d ≤ 0, H a : μ d > 0

Calculate the p -value: The p -value is 0.2150

Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p -value.

What is the conclusion?

At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. It should be noted that this data set consists of a very small sample size, with the possibility of an outlier data value. It is important to confirm that the matched pairs have differences that come from a population that is normal. The coach may want to consider increasing the sample size of players in the study.

Try It 10.12

A new prep class was designed to improve SAT test scores. Four students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in Table 10.18 . Are the scores, on average, higher after the class? Test at a 5% level.

Example 10.13

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in Table 10.19 .

Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.

Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation. x ¯ d x ¯ d = 3.71, s d s d = 4.5.

Random variable: X ¯ d X ¯ d = mean difference in the distances between the hands.

Distribution for the hypothesis test: t 6

H 0 : μ d = 0  H a : μ d ≠ 0

Calculate the p -value: The p -value is 0.0716 (using the data directly).

(test statistic = 2.18. p -value = 0.0719 using ( x ¯ d = 3.71 ,   s d = 4.5. ) ( x ¯ d = 3.71 ,   s d = 4.5. )

Decision: Assume α = 0.05. Since α < p -value, Do not reject H 0 .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put.

Try It 10.13

Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.20 . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.

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10.4 Matched or Paired Samples

DEPENDENT SAMPLES

If the choice of a sample member in one group automatically determines who’s going to be selected from the second group, then the samples are dependent. In these samples, there’s a distinct connection between members of the two samples. For example, when samples are taken from the same group at two different points in time (such as pre and post results) or under two different conditions (such as testing alone vs with a proctor), the two samples will be dependent. The group members can be different people and yet result in dependent samples if there is a natural connection between sample members from those two groups such as spouse/partner, twins, or some predefined criteria (gender, lifestyle, age, etc.) that can be used to match a member from one sample to another.

The requirements for hypothesis testing with dependent samples are:

• Samples taken are simple random samples with observations that are independent within each group. Sample sizes are often small.
• The two measurements (samples) are drawn from the same or same pair of individuals or objects (population variances should be the same)
• Either the matched pairs have differences that come from a population that is normal or the sampled number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.

If the above conditions hold, then we can use the sampling distribution of the mean of the differences between matched pairs in the sample, [latex]\bar d[/latex], to conduct our hypothesis test. Since the differences form just one list of values, we’ll perform a t-test. The test statistic for the dependent t -test is \[t = \frac{\bar d}{s_{\bar d}}\]

where [latex]\bar d[/latex] is the mean difference and [latex]s_{\bar d}[/latex] is the standard error of the mean difference, which is

\[s_{\bar d} = {s_d\over\sqrt{n}}\]where [latex]s_d[/latex] is the standard deviation of the differences of the data values from [latex]n[/latex] matched pairs in the samples.

Symbols Used in Hypothesis Testing for Means

Here’s an overview of testing with dependent samples.

WORKED OUT EXAMPLE – Dependent Samples

A local group of economists wants to study the effects of the pandemic on the small businesses in their city. A survey was sent out to a random sample of small businesses to anonymously report their weekly earnings for January of 2020 and a year later in January 2021. Assume that the weekly earnings are normally distributed. Only 10 businesses provided enough data as shown below. Data values are in thousands.

Use [latex]\alpha = 0.01[/latex] to decide whether there is sufficient evidence to support that the pandemic reduced average weekly earnings for small businesses.

We might be tempted to compare the average earnings of all of the small businesses in January 2020 with the average of all small businesses in 2021, that is, to test if the difference of averages is [latex]\mu_1 on average these differences will amount to zero , which is to say that there’s really not a whole lot of difference between the two groups.  The average of the differences from our samples is [latex]\bar d[/latex]. Let’s find that next by first finding the difference between January 2020 and January 2021 earnings for each pair:

The null hypothesis will be that on average these differences will amount to zero , which is to say that there’s really not a whole lot of difference between the two groups.  The average of the differences from our samples is [latex]\bar d[/latex]. Let’s find that next by first finding the difference between January 2020 and January 2021 earnings for each pair:

For the differences we can calculate their average and standard deviation by using a calculator (One Variable Statistics calculator at SUBEDI Calcs, for example).

The average of the differences, [latex]\bar d=6.48[/latex]. This is our sample statistic.

The standard deviation of the differences, [latex]s_d = 13.977028773431545[/latex].

In order to understand how usual or unusual this sample is, we need to know the sampling distribution of [latex]\bar d[/latex]. This [latex]\bar d[/latex] will be an estimate for the actual difference between the two populations, [latex]d[/latex]. On average, if we expect the January 2021 earnings to be lower, then we’d expect the average of the differences to be positive (if we compute differences between 2020 and 2021 for the same month of January: 2020  Earnings – 2021 Earnings . So the claim we would like to test is if, on average, the difference of January 2020 and January 2021 earnings is positive . This claim in symbols is: [latex]\mu_d > 0[/latex]. The opposite of this claim is [latex]\mu_d\le 0[/latex].

Since the differences are just one data list, we will use one sample t -test with data option to perform a t -test with the differences between the samples data calculated earlier (last row in the table above). However, if we’re using one of the calculators below, we can actually skip the step to calculate the differences. We only have to enter the data from each samples into the appropriate input boxes and enter a few other details about the test so that calculator can give us the correct results.

Sample information Sample data is paired difference in the sample.

STEP 1: Write the claim (or what’s being tested) using mathematical symbols On average the differences are positive. That is, [latex]\mu_d > 0[/latex].

STEP 2: Write the opposite of the claim using mathematical symbols Opposite of the claim: [latex]\mu_d\le 0[/latex].

STEP 3: Write the null and alternative hypothesis The null hypothesis is the statement that contains the condition of equality which in this case is the opposite of the claim

STEP 4: Identify the tail-type of the test and the level of significance, [latex]\alpha[/latex]. If [latex]\alpha[/latex] is not given, assume it to be 5% or 0.05. This is a right tailed test. Why? Since the level of significance is given, use that value of [latex]\alpha=0.01[/latex] (or 1% level of significance)

STEP 5: Check if CLT conditions are valid. The number of paired sample is small, lower than 30, so we need the paired differences in the population to be normally distributed to meet CLT conditions. We’ll perform a t-test for dependent means since the population standard deviation of the paired differences in the population is not known. This is the stage where we assume the equality in the null hypothesis is true, that is, assume [latex]\mu_d=0[/latex]. We’re assuming that there’s no change between the weekly earnings for small businesses in January of 2020 and January of 2021. We know that the sampling distribution of the mean of the paired difference, [latex]\bar d[/latex], will be approximately normally distributed with mean of [latex]\mu_{\bar d} = 0[/latex] and a standard deviation of [latex]\sigma_{\bar d} = \dfrac{\sigma_d}{\sqrt n}[/latex]. Since the population standard deviation of [latex]\sigma_d[/latex] is unknown, the standard deviation of the sampling distribution of [latex]\bar d[/latex] can be approximated as [latex]\sigma_{\bar d} \approx s_{\bar d}=\dfrac{s_d}{\sqrt n}[/latex] [1] . This sampling distribution can be approximated by a t -distribution with [latex]n-1[/latex] degrees of freedom, where  n is the number of matched pairs.

Results from STEP 6 can be obtained using a calculator. See below.

STEP 6: Plot t -score and find the p -value. Degrees of freedom for t -test: [latex]df = n - 1 = 10 - 1 = 9[/latex].

Use one or more of the calculators below to calculate p -value.

Go to Inference for the Mean @ rsubedi.com

Number of Samples

Independent or dependent samples?

Distribution Type for Test

Alternative Hypothesis

Confidence Interval? ← Optional

Enter Sample Data Enter your data for the two samples in the spreadsheet columns shown for data entry. You do not need to enter the difference data as the calculator will automatically calculate that to arrive at the test results. You can copy your data from the original source (comma or separated list, spreadsheet data, etc.) and paste that into the table on the calculator.

CALCULATE Results show in a panel to the right. Test statistic t , p -value, and df are displayed. The spreadsheet displays an additional column for the pair-wise difference of the sample data

Go to: Two Dependent Samples With Data  from the list of online calculators

Enter the following values and press  Calculate .

CALCULATE Results displayed are:

STEP 7: Make the decision by comparing p -value with α p-value is 0 . 0 8 8 3 3 5 7 4 4 8 3 5 8 which is more than [latex]\alpha[/latex] of 1%. Since p -value is more than [latex]\alpha[/latex], we fail to reject the null hypothesis ⇒ There’s not enough evidence to support Ha.

STEP 8: Interpret your decision in the context of original claim There is insufficient evidence at a significance level of 1% to support that the pandemic reduced average weekly earnings for small businesses.

• FSCJ Guides ↵

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8.1 Inference for Two Dependent Samples (Matched Pairs)

Learning Objectives

By the end of this chapter, the student should be able to:

• Classify hypothesis tests by type
• Conduct and interpret hypothesis tests for two population means, population standard deviations known
• Conduct and interpret hypothesis tests for two population means, population standard deviations unknown
• Conduct and interpret hypothesis tests for matched or paired samples
• Conduct and interpret hypothesis tests for two population proportions

Studies often compare two groups. For example, maybe researchers are interested in the effect aspirin has in preventing heart attacks.  One group is given aspirin and the other a placebo , and the heart attack rate is studied over several years.  Other studies may compare various diet and exercise programs.  Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether SAT or GRE preparatory courses really help raise their scores.

You have learned to conduct inference on single means and single proportions .  We know that the first step is deciding what type of data we are working with.  For quantitative data we are focused on means, while for categorical we are focused on proportions.  In this chapter we will compare two means or two proportions to each other.  The general procedure is still the same, just expanded.  With two sample analysis it is good to know what the formulas look like and where they come from, however you will probably lean heavily on technology in preforming the calculations.  

To compare two means we are obviously working with two groups, but first we need to think about the relationship between them. The groups are classified either as independent or dependent.  I ndependent samples consist of two samples that have no relationship, that is, sample values selected from one population are not related in any way to sample values selected from the other population.  Dependent samples consist of two groups that have some sort of identifiable relationship.

Two Dependent Samples (Matched Pairs)

Two samples that are dependent typically come from a matched pairs experimental design. The parameter tested using matched pairs is the population mean difference .  When using inference techniques for matched or paired samples, the following characteristics should be present:

• Simple random sampling is used.
• Sample sizes are often small.
• Two measurements (samples) are drawn from the same pair of (or two extremely similar) individuals or objects.
• Differences are calculated from the matched or paired samples.
• The differences form the sample that is used for analysis.

Confidence intervals may be calculated on their own for two samples but often, especially in the case of matched pairs, we first want to formally check to see if a difference exists with a hypothesis test.  If we do find a statistically significant difference then we may estimate it with a CI after the fact.

Hypothesis Tests for the Mean difference

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated, and the population mean difference, μ d , is our parameter of interest.  Although it is possible to test for a certain magnitude of effect, we are most often just looking for a general effect.  Our hypothesis would then look like:

H o : μ d =0

H a : μ d (<, >, ≠) 0

The steps are the same as we are familiar with, but it is tested using a Student’s-t test for a single population mean with n – 1 degrees of freedom, with the test statistic:

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the figure below. A lower score indicates less pain. The “before” value is matched to an “after” value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level.

Confidence Intervals for the Mean difference

If we are using the t distribution, the error bound for the population mean difference is:

• use df = n – 1 degrees of freedom, where n is the number of pairs
• s d =  standard deviation of the differences.

A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:

The coach wants to know if the strength development class makes his players stronger, on average.

Using the differences data, calculate the sample mean and the sample standard deviation.

Using the difference data, this becomes a test of a single __________ (fill in the blank).

Calculate the p -value:

What is the conclusion?

A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in the figure below. Are the scores, on average, higher after the class? Test at a 5% level.

Image Credits

Figure 8.1: Ali Inay (2015). “Brunching with Friends.” Public domain. Retrieved from https://unsplash.com/photos/y3aP9oo9Pjc

Figure 8.3: Kindred Grey via Virginia Tech (2020). “Figure 8.3” CC BY-SA 4.0. Retrieved from https://commons.wikimedia.org/wiki/File:Figure_8.3.png . Adaptation of Figure 5.39 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/statistics/pages/5-practice

Figure 8.6: Kindred Grey via Virginia Tech (2020). “Figure 8.6” CC BY-SA 4.0. Retrieved from https://commons.wikimedia.org/wiki/File:Figure_8.6.png . Adaptation of Figure 5.39 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/statistics/pages/5-practice

An inactive treatment that has no real effect on the explanatory variable

The facet of statistics dealing with using a sample to generalize (or infer) about the population

The arithmetic mean, or average of a population

The number of individuals that have a characteristic we are interested in divided by the total number in the population

Numerical data with a mathematical context

Data that describes qualities, or puts individuals into categories

The occurrence of one event has no effect on the probability of the occurrence of another event

Very similar individuals (or even the same individual) receive two different two treatments (or treatment vs. control) then the difference in results are compared

The mean of the differences in a matched pairs design

The probability distribution of a statistic at a given sample size

The value that is calculated from a sample used to estimate an unknown population parameter

Significant Statistics Copyright © 2020 by John Morgan Russell, OpenStaxCollege, OpenIntro is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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BUS204: Business Statistics

Hypothesis Testing with Two Samples

Read this chapter, which discusses how to compare data from two similar groups. This is useful when, for example, you want to analyze things like how someone's income relates to another sample that you are interested in. Make sure you read the introduction as well as sections 10.1 through 10.6. Attempt the practice problems and homework at the end of the chapter.

Matched or Paired Samples

• Simple random sampling is used.
• Sample sizes are often small.
• Two measurements (samples) are drawn from the same pair of individuals or objects.
• Differences are calculated from the matched or paired samples.
• The differences form the sample that is used for the hypothesis test.
• Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences, that is, the number of pairs not the number of observations.

The null and alternative hypotheses for this test are:

The test statistic is:

Example 10.9

Problem A company has developed a training program for its entering employees because they have become concerned with the results of the six-month employee review. They hope that the training program can result in better six-month reviews. Each trainee constitutes a "pair", the entering score the employee received when first entering the firm and the score given at the six-month review. The difference in the two scores were calculated for each employee and the means for before and after the training program was calculated. The sample mean before the training program was 20.4 and the sample mean after the training program was 23.9. The standard deviation of the differences in the two scores across the 20 employees was 3.8 points. Test at the 10% significance level the null hypothesis that the two population means are equal against the alternative that the training program helps improve the employees' scores.

Solution 1 The first step is to identify this as a two sample case: before the training and after the training. This differentiates this problem from simple one sample issues. Second, we determine that the two samples are "paired". Each observation in the first sample has a paired observation in the second sample. This information tells us that the null and alternative hypotheses should be:

This form reflects the implied claim that the training course improves scores; the test is one-tailed and the claim is in the alternative hypothesis. Because the experiment was conducted as a matched paired sample rather than simply taking scores from people who took the training course those who didn't, we use the matched pair test statistic:

In order to solve this equation, the individual scores, pre-training course and post-training course need to be used to calculate the individual differences. These scores are then averaged and the average difference is calculated:

From these differences we can calculate the standard deviation across the individual differences:

We can now compare the calculated value of the test statistic, 4.12, with the critical value. The critical value is a Student's t with degrees of freedom equal to the number of pairs, not observations, minus 1. In this case 20 pairs and at 90% confidence level t a/2 = ±1.729 at df = 20 - 1 = 19. The calculated test statistic is most certainly in the tail of the distribution and thus we cannot accept the null hypothesis that there is no difference from the training program. Evidence seems indicate that the training aids employees in gaining higher scores.

Example 10.10

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.5. A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Example 10.11

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Chapter 10.5: Matched or Paired Samples

When using a hypothesis test for matched or paired samples, the following characteristics should be present:

• Simple random sampling is used.
• Sample sizes are often small.
• Two measurements (samples) are drawn from the same pair of individuals or objects.
• Differences are calculated from the matched or paired samples.
• The differences form the sample that is used for the hypothesis test.
• Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.

In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μ d , is then tested using a Student’s-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences.

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in (Figure) . A lower score indicates less pain. The “before” value is matched to an “after” value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Corresponding “before” and “after” values form matched pairs. (Calculate “after” – “before.”)

The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6}

H 0 : μ d ≥ 0

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μ d is the population mean of the differences.)

H a : μ d < 0

The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student’s t with df = n – 1 = 8 – 1 = 7. Use t 7 . (Notice that the test is for a single population mean.)

Calculate the p -value using the Student’s-t distribution: p -value = 0.0095

The sample mean and sample standard deviation of the differences are:

Compare α and the p -value: α = 0.05 and p -value = 0.0095. α > p -value.

Make a decision: Since α > p -value, reject H 0 . This means that μ d < 0 and there is improvement.

Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.

For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time ( after – before ) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1 st list name – 2 nd list name. The calculator will do the subtraction, and you will have the differences in the third list.

A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level.

A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:

The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation.

The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative.

Using the difference data, this becomes a test of a single __________ (fill in the blank).

The distribution for the hypothesis test is t 3 .

H 0 : μ d ≤ 0, H a : μ d > 0

Calculate the p -value: The p -value is 0.2150

Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p -value.

What is the conclusion?

At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.

A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in (Figure) . Are the scores, on average, higher after the class? Test at a 5% level.

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in (Figure) .

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Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.

Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution.

Distribution for the hypothesis test: t 6

H 0 : μ d = 0  H a : μ d ≠ 0

Calculate the p -value: The p -value is 0.0716 (using the data directly).

Decision: Assume α = 0.05. Since α < p -value, Do not reject H 0 .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put.

Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in (Figure) . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.

Chapter Review

A hypothesis test for matched or paired samples (t-test) has these characteristics:

• Test the differences by subtracting one measurement from the other measurement
• Distribution: Student’s-t distribution with n – 1 degrees of freedom
• If the number of differences is small (less than 30), the differences must follow a normal distribution.
• Two samples are drawn from the same set of objects.
• Samples are dependent.

Formula Review

Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in (Figure) . The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.

What is the random variable?

the mean difference of the system failures

State the null and alternative hypotheses.

What is the p -value?

Draw the graph of the p -value.

What conclusion can you draw about the software patch?

With a p -value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level.

What is the sample mean difference?

What conclusion can you draw about the juggling class?

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.

What is the test statistic?

We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.

DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student’s t -distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.)

1) Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in (Figure) . Do you think that their cholesterol levels were significantly lowered?

p -value = 0.1494

Use the following information to answer the next two exercises.

A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same.

Let the subscript t = treated patient and ut = untreated patient.

2) The appropriate hypotheses are:

• H 0 : p t < p ut and H a : p t ≥ p ut
• H 0 : p t ≤ p ut and H a : p t > p ut
• H 0 : p t = p ut and H a : p t ≠ p ut
• H 0 : p t = p ut and H a : p t < p ut

3) If the p -value is 0.0062 what is the conclusion (use α = 0.05)?

• The method has no effect.
• There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.
• There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients who develop AIDS after four years.
• There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years.

4) The distribution for the test is:

• N (−10.2, 8.4)

5) If α = 0.05, the p -value and the conclusion are

• 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training.
• 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training.
• 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training.
• 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training.

6) A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows.

The correct decision is:

• Reject H 0 .
• Do not reject the H 0 .

7) A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in (Figure) .

8) A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in (Figure) . Test at the 1% level of significance.

9) A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in (Figure) .

Bringing It Together

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test.

• independent group means, population standard deviations and/or variances known
• independent group means, population standard deviations and/or variances unknown
• matched or paired samples
• single mean
• two proportions
• single proportion

10) A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.

11) A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

12) The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females.

13) A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.

14) A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.

15) According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this.

16) According to a recent study, U.S. companies have a mean maternity-leave of six weeks.

17) A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally.

18) A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected:

19) University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.

20) Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown (Figure) . Determine the appropriate test and best distribution to use for that test.

• Two independent means, normal distribution
• Two independent means, Student’s-t distribution
• Matched or paired samples, Student’s-t distribution
• Two population proportions, normal distribution

21) A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as (Figure) .

• a test of two independent means.
• a test of two proportions.
• a test of a single mean.
• a test of a single proportion.

Answers to odd questions

1) At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks.

Test: two matched pairs or paired samples ( t -test)

Distribution: t 12

H 0 : μ d = 0 H a : μ d > 0

The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012.

Graph: right-tailed

p -value: 0.0004

Decision: Reject H 0

Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012.

Test: matched or paired samples ( t -test)

Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, 1.8}

Distribution: H 0 : μ d = 0 H a : μ d < 0

The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012.

Graph: left-tailed.

p -value: 0.1207

Decision: Do not reject H 0 .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012.

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