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9.4: Distribution Needed for Hypothesis Testing

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Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's $t$-distribution. (Remember, use a Student's $t$-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually $n$ is large or the sample size is large).

If you are testing a single population mean, the distribution for the test is for means :

\[\bar{X} - N\left(\mu_{x}, \frac{\sigma_{x}}{\sqrt{n}}\right)\]

The population parameter is $\mu$. The estimated value (point estimate) for $\mu$ is $\bar{x}$, the sample mean.

If you are testing a single population proportion, the distribution for the test is for proportions or percentages:

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\]

The population parameter is $p$. The estimated value (point estimate) for $p$ is $p′$. $p' = \frac{x}{n}$ where $x$ is the number of successes and n is the sample size.

Assumptions

When you perform a hypothesis test of a single population mean $\mu$ using a Student's $t$-distribution (often called a $t$-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed. You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a $t$-test will work even if the population is not approximately normally distributed).

When you perform a hypothesis test of a single population mean $\mu$ using a normal distribution (often called a $z$-test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.

When you perform a hypothesis test of a single population proportion $p$, you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are: there are a certain number $n$ of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success $p$. The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities $np$ and $nq$ must both be greater than five $(np > 5$ and $nq > 5)$. Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with $\mu = p$ and $\sigma = \sqrt{\frac{pq}{n}}$. Remember that $q = 1 – p$.

In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.

When testing for a single population mean:

A Student's $t$-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation.

When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of successes and the mean number of failures satisfy the conditions: $np > 5$ and $nq > 5$ where $n$ is the sample size, $p$ is the probability of a success, and $q$ is the probability of a failure.

Formula Review

If there is no given preconceived $\alpha$, then use $\alpha = 0.05$.

Types of Hypothesis Tests

Single population mean, known population variance (or standard deviation): Normal test .
Single population mean, unknown population variance (or standard deviation): Student's $t$-test .
Single population proportion: Normal test .
For a single population mean , we may use a normal distribution with the following mean and standard deviation. Means: $\mu = \mu_{\bar{x}}$ and $\\sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}}$
A single population proportion , we may use a normal distribution with the following mean and standard deviation. Proportions: $\mu = p$ and $\sigma = \sqrt{\frac{pq}{n}}$.
It is continuous and assumes any real values.
The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution.
It approaches the standard normal distribution as $n$ gets larger.
There is a "family" of $t$-distributions: every representative of the family is completely defined by the number of degrees of freedom which is one less than the number of data items.

9.3 Distribution Needed for Hypothesis Testing

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t -distribution . (Remember, use a Student's t -distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large).

Assumptions

When you perform a hypothesis test of a single population mean μ using a Student's t -distribution (often called a t -test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed . You use the sample standard deviation to approximate the population standard deviation. Note that if the sample size is sufficiently large, a t -test will work even if the population is not approximately normally distributed.

When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z -test), you take a simple random sample from the population. The population you are testing is normally distributed or your sample size is sufficiently large. You know the value of the population standard deviation which, in reality, is rarely known.

When you perform a hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution , which are the following: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and σ = p q n σ = p q n . Remember that q = 1 – p .

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Keyboard Shortcuts

Hypothesis testing.

Key Topics:

Basic approach
Null and alternative hypothesis
Decision making and the p -value
Z-test & Nonparametric alternative

Basic approach to hypothesis testing

State a model describing the relationship between the explanatory variables and the outcome variable(s) in the population and the nature of the variability. State all of your assumptions .
Specify the null and alternative hypotheses in terms of the parameters of the model.
Invent a test statistic that will tend to be different under the null and alternative hypotheses.
Using the assumptions of step 1, find the theoretical sampling distribution of the statistic under the null hypothesis of step 2. Ideally the form of the sampling distribution should be one of the “standard distributions”(e.g. normal, t , binomial..)
Calculate a p -value , as the area under the sampling distribution more extreme than your statistic. Depends on the form of the alternative hypothesis.
Choose your acceptable type 1 error rate (alpha) and apply the decision rule : reject the null hypothesis if the p-value is less than alpha, otherwise do not reject.
$\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}$
general form is: (estimate - value we are testing)/(st.dev of the estimate)
z-statistic follows N(0,1) distribution
2 × the area above |z|, area above z,or area below z, or
compare the statistic to a critical value, |z| ≥ z α/2 , z ≥ z α , or z ≤ - z α
Choose the acceptable level of Alpha = 0.05, we conclude …. ?

Making the Decision

It is either likely or unlikely that we would collect the evidence we did given the initial assumption. (Note: “likely” or “unlikely” is measured by calculating a probability!)

If it is likely , then we “ do not reject ” our initial assumption. There is not enough evidence to do otherwise.

If it is unlikely , then:

either our initial assumption is correct and we experienced an unusual event or,
our initial assumption is incorrect

In statistics, if it is unlikely, we decide to “ reject ” our initial assumption.

Example: Criminal Trial Analogy

First, state 2 hypotheses, the null hypothesis (“H 0 ”) and the alternative hypothesis (“H A ”)

H 0 : Defendant is not guilty.
H A : Defendant is guilty.

Usually the H 0 is a statement of “no effect”, or “no change”, or “chance only” about a population parameter.

While the H A , depending on the situation, is that there is a difference, trend, effect, or a relationship with respect to a population parameter.

It can one-sided and two-sided.
In two-sided we only care there is a difference, but not the direction of it. In one-sided we care about a particular direction of the relationship. We want to know if the value is strictly larger or smaller.

Then, collect evidence, such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, handwriting samples, etc. (In statistics, the data are the evidence.)

Next, you make your initial assumption.

Defendant is innocent until proven guilty.

In statistics, we always assume the null hypothesis is true .

Then, make a decision based on the available evidence.

If there is sufficient evidence (“beyond a reasonable doubt”), reject the null hypothesis . (Behave as if defendant is guilty.)
If there is not enough evidence, do not reject the null hypothesis . (Behave as if defendant is not guilty.)

If the observed outcome, e.g., a sample statistic, is surprising under the assumption that the null hypothesis is true, but more probable if the alternative is true, then this outcome is evidence against H 0 and in favor of H A .

An observed effect so large that it would rarely occur by chance is called statistically significant (i.e., not likely to happen by chance).

Using the p -value to make the decision

The p -value represents how likely we would be to observe such an extreme sample if the null hypothesis were true. The p -value is a probability computed assuming the null hypothesis is true, that the test statistic would take a value as extreme or more extreme than that actually observed. Since it's a probability, it is a number between 0 and 1. The closer the number is to 0 means the event is “unlikely.” So if p -value is “small,” (typically, less than 0.05), we can then reject the null hypothesis.

Significance level and p -value

Significance level, α, is a decisive value for p -value. In this context, significant does not mean “important”, but it means “not likely to happened just by chance”.

α is the maximum probability of rejecting the null hypothesis when the null hypothesis is true. If α = 1 we always reject the null, if α = 0 we never reject the null hypothesis. In articles, journals, etc… you may read: “The results were significant ( p <0.05).” So if p =0.03, it's significant at the level of α = 0.05 but not at the level of α = 0.01. If we reject the H 0 at the level of α = 0.05 (which corresponds to 95% CI), we are saying that if H 0 is true, the observed phenomenon would happen no more than 5% of the time (that is 1 in 20). If we choose to compare the p -value to α = 0.01, we are insisting on a stronger evidence!

So, what kind of error could we make? No matter what decision we make, there is always a chance we made an error.

Errors in Criminal Trial:

Errors in Hypothesis Testing

Type I error (False positive): The null hypothesis is rejected when it is true.

α is the maximum probability of making a Type I error.

Type II error (False negative): The null hypothesis is not rejected when it is false.

β is the probability of making a Type II error

There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!

The power of a statistical test is its probability of rejecting the null hypothesis if the null hypothesis is false. That is, power is the ability to correctly reject H 0 and detect a significant effect. In other words, power is one minus the type II error risk.

$\text{Power }=1-\beta = P\left(\text{reject} H_0 | H_0 \text{is false } \right)$

Which error is worse?

Type I = you are innocent, yet accused of cheating on the test. Type II = you cheated on the test, but you are found innocent.

This depends on the context of the problem too. But in most cases scientists are trying to be “conservative”; it's worse to make a spurious discovery than to fail to make a good one. Our goal it to increase the power of the test that is to minimize the length of the CI.

We need to keep in mind:

the effect of the sample size,
the correctness of the underlying assumptions about the population,
statistical vs. practical significance, etc…

(see the handout). To study the tradeoffs between the sample size, α, and Type II error we can use power and operating characteristic curves.

What type of error might we have made?

Type I error is claiming that average student height is not 65 inches, when it really is. Type II error is failing to claim that the average student height is not 65in when it is.

We rejected the null hypothesis, i.e., claimed that the height is not 65, thus making potentially a Type I error. But sometimes the p -value is too low because of the large sample size, and we may have statistical significance but not really practical significance! That's why most statisticians are much more comfortable with using CI than tests.

There is a need for a further generalization. What if we can't assume that σ is known? In this case we would use s (the sample standard deviation) to estimate σ.

If the sample is very large, we can treat σ as known by assuming that σ = s . According to the law of large numbers, this is not too bad a thing to do. But if the sample is small, the fact that we have to estimate both the standard deviation and the mean adds extra uncertainty to our inference. In practice this means that we need a larger multiplier for the standard error.

We need one-sample t -test.

One sample t -test

Assume data are independently sampled from a normal distribution with unknown mean μ and variance σ 2 . Make an initial assumption, μ 0 .
t-statistic: $\frac{\bar{X}-\mu_0}{s / \sqrt{n}}$ where s is a sample st.dev.
t-statistic follows t -distribution with df = n - 1
Alpha = 0.05, we conclude ….

Testing for the population proportion

Let's go back to our CNN poll. Assume we have a SRS of 1,017 adults.

We are interested in testing the following hypothesis: H 0 : p = 0.50 vs. p > 0.50

What is the test statistic?

If alpha = 0.05, what do we conclude?

We will see more details in the next lesson on proportions, then distributions, and possible tests.

Module 9: Hypothesis Testing With One Sample

Distribution needed for hypothesis testing, learning outcomes.

Conduct and interpret hypothesis tests for a single population mean, population standard deviation known
Conduct and interpret hypothesis tests for a single population mean, population standard deviation unknown

Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student’s t- distribution . (Remember, use a Student’s t -distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.) We perform tests of a population proportion using a normal distribution (usually n is large or the sample size is large).

If you are testing a single population mean , the distribution for the test is for means :

[latex]\displaystyle\overline{{X}}\text{~}{N}{\left(\mu_{{X}}\text{ , }\frac{{\sigma_{{X}}}}{\sqrt{{n}}}\right)}{\quad\text{or}\quad}{t}_{{{d}{f}}}[/latex]

The population parameter is [latex]\mu[/latex]. The estimated value (point estimate) for [latex]\mu[/latex] is [latex]\displaystyle\overline{{x}}[/latex], the sample mean.

If you are testing a single population proportion , the distribution for the test is for proportions or percentages:

[latex]\displaystyle{P}^{\prime}\text{~}{N}{\left({p}\text{ , }\sqrt{{\frac{{{p}{q}}}{{n}}}}\right)}[/latex]

The population parameter is [latex]p[/latex]. The estimated value (point estimate) for [latex]p[/latex] is p′ . [latex]\displaystyle{p}\prime=\frac{{x}}{{n}}[/latex] where [latex]x[/latex] is the number of successes and [latex]n[/latex] is the sample size.

Assumptions

When you perform a hypothesis test of a single population mean μ using a Student’s t -distribution (often called a t-test), there are fundamental assumptions that need to be met in order for the test to work properly. Your data should be a simple random sample that comes from a population that is approximately normally distributed . You use the sample standard deviation to approximate the population standard deviation. (Note that if the sample size is sufficiently large, a t-test will work even if the population is not approximately normally distributed).

When you perform a hypothesis test of a single population proportion p , you take a simple random sample from the population. You must meet the conditions for a binomial distribution which are as follows: there are a certain number n of independent trials, the outcomes of any trial are success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five ( np > 5 and nq > 5). Then the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = p and [latex]\displaystyle\sigma=\sqrt{{\frac{{{p}{q}}}{{n}}}}[/latex] . Remember that q = 1 – p .

Concept Review

In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied.

When testing for a single population mean:

A Student’s t -test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation.
The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation.

When testing a single population proportion use a normal test for a single population proportion if the data comes from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure.

Formula Review

If there is no given preconceived α , then use α = 0.05.

Types of Hypothesis Tests

Single population mean, known population variance (or standard deviation): Normal test .
Single population mean, unknown population variance (or standard deviation): Student’s t -test .
Single population proportion: Normal test .
For a single population mean , we may use a normal distribution with the following mean and standard deviation. Means: [latex]\displaystyle\mu=\mu_{{\overline{{x}}}}{\quad\text{and}\quad}\sigma_{{\overline{{x}}}}=\frac{{\sigma_{{x}}}}{\sqrt{{n}}}[/latex]
A single population proportion , we may use a normal distribution with the following mean and standard deviation. Proportions: [latex]\displaystyle\mu={p}{\quad\text{and}\quad}\sigma=\sqrt{{\frac{{{p}{q}}}{{n}}}}[/latex].
Distribution Needed for Hypothesis Testing. Provided by : OpenStax. Located at : . License : CC BY: Attribution
Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

An Introduction to Bayesian Thinking

Chapter 5 hypothesis testing with normal populations.

In Section 3.5 , we described how the Bayes factors can be used for hypothesis testing. Now we will use the Bayes factors to compare normal means, i.e., test whether the mean of a population is zero or compare the means of two groups of normally-distributed populations. We divide this mission into three cases: known variance for a single population, unknown variance for a single population using paired data, and unknown variance using two independent groups.

Also note that some of the examples in this section use an updated version of the bayes_inference function. If your local output is different from what is seen in this chapter, or the provided code fails to run for you please make sure that you have the most recent version of the package.

5.1 Bayes Factors for Testing a Normal Mean: variance known

Now we show how to obtain Bayes factors for testing hypothesis about a normal mean, where the variance is known . To start, let’s consider a random sample of observations from a normal population with mean $\mu$ and pre-specified variance $\sigma^2$ . We consider testing whether the population mean $\mu$ is equal to $m_0$ or not.

Therefore, we can formulate the data and hypotheses as below:

Data \[Y_1, \cdots, Y_n \mathrel{\mathop{\sim}\limits^{\rm iid}}\textsf{Normal}(\mu, \sigma^2)\]

$H_1: \mu = m_0$
$H_2: \mu \neq m_0$

We also need to specify priors for $\mu$ under both hypotheses. Under $H_1$ , we assume that $\mu$ is exactly $m_0$ , so this occurs with probability 1 under $H_1$ . Now under $H_2$ , $\mu$ is unspecified, so we describe our prior uncertainty with the conjugate normal distribution centered at $m_0$ and with a variance $\sigma^2/\mathbf{n_0}$ . This is centered at the hypothesized value $m_0$ , and it seems that the mean is equally likely to be larger or smaller than $m_0$ , so a dividing factor $n_0$ is given to the variance. The hyper parameter $n_0$ controls the precision of the prior as before.

In mathematical terms, the priors are:

$H_1: \mu = m_0 \text{ with probability 1}$
$H_2: \mu \sim \textsf{Normal}(m_0, \sigma^2/\mathbf{n_0})$

Bayes Factor

Now the Bayes factor for comparing $H_1$ to $H_2$ is the ratio of the distribution, the data under the assumption that $\mu = m_0$ to the distribution of the data under $H_2$ .

\[\begin{aligned} \textit{BF}[H_1 : H_2] &= \frac{p(\text{data}\mid \mu = m_0, \sigma^2 )} {\int p(\text{data}\mid \mu, \sigma^2) p(\mu \mid m_0, \mathbf{n_0}, \sigma^2)\, d \mu} \\ \textit{BF}[H_1 : H_2] &=\left(\frac{n + \mathbf{n_0}}{\mathbf{n_0}} \right)^{1/2} \exp\left\{-\frac 1 2 \frac{n }{n + \mathbf{n_0}} Z^2 \right\} \\ Z &= \frac{(\bar{Y} - m_0)}{\sigma/\sqrt{n}} \end{aligned}\]

The term in the denominator requires integration to account for the uncertainty in $\mu$ under $H_2$ . And it can be shown that the Bayes factor is a function of the observed sampled size, the prior sample size $n_0$ and a $Z$ score.

Let’s explore how the hyperparameters in $n_0$ influences the Bayes factor in Equation (5.1) . For illustration we will use the sample size of 100. Recall that for estimation, we interpreted $n_0$ as a prior sample size and considered the limiting case where $n_0$ goes to zero as a non-informative or reference prior.

\[\begin{equation} \textsf{BF}[H_1 : H_2] = \left(\frac{n + \mathbf{n_0}}{\mathbf{n_0}}\right)^{1/2} \exp\left\{-\frac{1}{2} \frac{n }{n + \mathbf{n_0}} Z^2 \right\} \tag{5.1} \end{equation}\]

Figure 5.1 shows the Bayes factor for comparing $H_1$ to $H_2$ on the y-axis as $n_0$ changes on the x-axis. The different lines correspond to different values of the $Z$ score or how many standard errors $\bar{y}$ is from the hypothesized mean. As expected, larger values of the $Z$ score favor $H_2$ .

Figure 5.1: Vague prior for mu: n=100

But as $n_0$ becomes smaller and approaches 0, the first term in the Bayes factor goes to infinity, while the exponential term involving the data goes to a constant and is ignored. In the limit as $n_0 \rightarrow 0$ under this noninformative prior, the Bayes factor paradoxically ends up favoring $H_1$ regardless of the value of $\bar{y}$ .

The takeaway from this is that we cannot use improper priors with $n_0 = 0$ , if we are going to test our hypothesis that $\mu = n_0$ . Similarly, vague priors that use a small value of $n_0$ are not recommended due to the sensitivity of the results to the choice of an arbitrarily small value of $n_0$ .

This problem arises with vague priors – the Bayes factor favors the null model $H_1$ even when the data are far away from the value under the null – are known as the Bartlett’s paradox or the Jeffrey’s-Lindleys paradox.

Now, one way to understand the effect of prior is through the standard effect size

\[\delta = \frac{\mu - m_0}{\sigma}.\] The prior of the standard effect size is

\[\delta \mid H_2 \sim \textsf{Normal}(0, \frac{1}{\mathbf{n_0}})\]

This allows us to think about a standardized effect independent of the units of the problem. One default choice is using the unit information prior, where the prior sample size $n_0$ is 1, leading to a standard normal for the standardized effect size. This is depicted with the blue normal density in Figure 5.2 . This suggested that we expect that the mean will be within $\pm 1.96$ standard deviations of the hypothesized mean with probability 0.95 . (Note that we can say this only under a Bayesian setting.)

In many fields we expect that the effect will be small relative to $\sigma$ . If we do not expect to see large effects, then we may want to use a more informative prior on the effect size as the density in orange with $n_0 = 4$ . So they expected the mean to be within $\pm 1/\sqrt{n_0}$ or five standard deviations of the prior mean.

Figure 5.2: Prior on standard effect size

Example 1.1 To illustrate, we give an example from parapsychological research. The case involved the test of the subject’s claim to affect a series of randomly generated 0’s and 1’s by means of extra sensory perception (ESP). The random sequence of 0’s and 1’s are generated by a machine with probability of generating 1 being 0.5. The subject claims that his ESP would make the sample mean differ significantly from 0.5.

Therefore, we are testing $H_1: \mu = 0.5$ versus $H_2: \mu \neq 0.5$ . Let’s use a prior that suggests we do not expect a large effect which leads the following solution for $n_0$ . Assume we want a standard effect of 0.03, there is a 95% chance that it is between $(-0.03/\sigma, 0.03/\sigma)$ , with $n_0 = (1.96\sigma/0.03)^2 = 32.7^2$ .

Figure 5.3 shows our informative prior in blue, while the unit information prior is in orange. On this scale, the unit information prior needs to be almost uniform for the range that we are interested.

Figure 5.3: Prior effect in the extra sensory perception test

A very large data set with over 104 million trials was collected to test this hypothesis, so we use a normal distribution to approximate the distribution the sample mean.

Sample size: $n = 1.0449 \times 10^8$
Sample mean: $\bar{y} = 0.500177$ , standard deviation $\sigma = 0.5$
$Z$ -score: 3.61

Now using our prior in the data, the Bayes factor for $H_1$ to $H_2$ was 0.46, implying evidence against the hypothesis $H_1$ that $\mu = 0.5$ .

Informative $\textit{BF}[H_1:H_2] = 0.46$
$\textit{BF}[H_2:H_1] = 1/\textit{BF}[H_1:H_2] = 2.19$

Now, this can be inverted to provide the evidence in favor of $H_2$ . The evidence suggests that the hypothesis that the machine operates with a probability that is not 0.5, is 2.19 times more likely than the hypothesis the probability is 0.5. Based on the interpretation of Bayes factors from Table 3.5 , this is in the range of “not worth the bare mention”.

To recap, we present expressions for calculating Bayes factors for a normal model with a specified variance. We show that the improper reference priors for $\mu$ when $n_0 = 0$ , or vague priors where $n_0$ is arbitrarily small, lead to Bayes factors that favor the null hypothesis regardless of the data, and thus should not be used for hypothesis testing.

Bayes factors with normal priors can be sensitive to the choice of the $n_0$ . While the default value of $n_0 = 1$ is reasonable in many cases, this may be too non-informative if one expects more effects. Wherever possible, think about how large an effect you expect and use that information to help select the $n_0$ .

All the ESP examples suggest weak evidence and favored the machine generating random 0’s and 1’s with a probability that is different from 0.5. Note that ESP is not the only explanation – a deviation from 0.5 can also occur if the random number generator is biased. Bias in the stream of random numbers in our pseudorandom numbers has huge implications for numerous fields that depend on simulation. If the context had been about detecting a small bias in random numbers what prior would you use and how would it change the outcome? You can experiment it in R or other software packages that generate random Bernoulli trials.

Next, we will look at Bayes factors in normal models with unknown variances using the Cauchy prior so that results are less sensitive to the choice of $n_0$ .

5.2 Comparing Two Paired Means using Bayes Factors

We previously learned that we can use a paired t-test to compare means from two paired samples. In this section, we will show how Bayes factors can be expressed as a function of the t-statistic for comparing the means and provide posterior probabilities of the hypothesis that whether the means are equal or different.

Example 5.1 Trace metals in drinking water affect the flavor, and unusually high concentrations can pose a health hazard. Ten pairs of data were taken measuring the zinc concentration in bottom and surface water at ten randomly sampled locations, as listed in Table 5.1 .

Water samples collected at the the same location, on the surface and the bottom, cannot be assumed to be independent of each other. However, it may be reasonable to assume that the differences in the concentration at the bottom and the surface in randomly sampled locations are independent of each other.

To start modeling, we will treat the ten differences as a random sample from a normal population where the parameter of interest is the difference between the average zinc concentration at the bottom and the average zinc concentration at the surface, or the main difference, $\mu$ .

In mathematical terms, we have

Random sample of $n= 10$ differences $Y_1, \ldots, Y_n$
Normal population with mean $\mu \equiv \mu_B - \mu_S$

In this case, we have no information about the variability in the data, and we will treat the variance, $\sigma^2$ , as unknown.

The hypothesis of the main concentration at the surface and bottom are the same is equivalent to saying $\mu = 0$ . The second hypothesis is that the difference between the mean bottom and surface concentrations, or equivalently that the mean difference $\mu \neq 0$ .

In other words, we are going to compare the following hypotheses:

$H_1: \mu_B = \mu_S \Leftrightarrow \mu = 0$
$H_2: \mu_B \neq \mu_S \Leftrightarrow \mu \neq 0$

The Bayes factor is the ratio between the distributions of the data under each hypothesis, which does not depend on any unknown parameters.

\[\textit{BF}[H_1 : H_2] = \frac{p(\text{data}\mid H_1)} {p(\text{data}\mid H_2)}\]

To obtain the Bayes factor, we need to use integration over the prior distributions under each hypothesis to obtain those distributions of the data.

\[\textit{BF}[H_1 : H_2] = \iint p(\text{data}\mid \mu, \sigma^2) p(\mu \mid \sigma^2) p(\sigma^2 \mid H_2)\, d \mu \, d\sigma^2\]

This requires specifying the following priors:

$\mu \mid \sigma^2, H_2 \sim \textsf{Normal}(0, \sigma^2/n_0)$
$p(\sigma^2) \propto 1/\sigma^2$ for both $H_1$ and $H_2$

$\mu$ is exactly zero under the hypothesis $H_1$ . For $\mu$ in $H_2$ , we start with the same conjugate normal prior as we used in Section 5.1 – testing the normal mean with known variance. Since we assume that $\sigma^2$ is known, we model $\mu \mid \sigma^2$ instead of $\mu$ itself.

The $\sigma^2$ appears in both the numerator and denominator of the Bayes factor. For default or reference case, we use the Jeffreys prior (a.k.a. reference prior) on $\sigma^2$ . As long as we have more than two observations, this (improper) prior will lead to a proper posterior.

After integration and rearranging, one can derive a simple expression for the Bayes factor:

\[\textit{BF}[H_1 : H_2] = \left(\frac{n + n_0}{n_0} \right)^{1/2} \left( \frac{ t^2 \frac{n_0}{n + n_0} + \nu } { t^2 + \nu} \right)^{\frac{\nu + 1}{2}}\]

This is a function of the t-statistic

\[t = \frac{|\bar{Y}|}{s/\sqrt{n}},\]

where $s$ is the sample standard deviation and the degrees of freedom $\nu = n-1$ (sample size minus one).

As we saw in the case of Bayes factors with known variance, we cannot use the improper prior on $\mu$ because when $n_0 \to 0$ , then $\textit{BF}[H1:H_2] \to \infty$ favoring $H_1$ regardless of the magnitude of the t-statistic. Arbitrary, vague small choices for $n_0$ also lead to arbitrary large Bayes factors in favor of $H_1$ . Another example of the Barlett’s or Jeffreys-Lindley paradox.

Sir Herald Jeffrey discovered another paradox testing using the conjugant normal prior, known as the information paradox . His thought experiment assumed that our sample size $n$ and the prior sample size $n_0$ . He then considered what would happen to the Bayes factor as the sample mean moved further and further away from the hypothesized mean, measured in terms standard errors with the t-statistic, i.e., $|t| \to \infty$ . As the t-statistic or information about the mean moved further and further from zero, the Bayes factor goes to a constant depending on $n, n_0$ rather than providing overwhelming support for $H_2$ .

The bounded Bayes factor is

\[\textit{BF}[H_1 : H_2] \to \left( \frac{n_0}{n_0 + n} \right)^{\frac{n - 1}{2}}\]

Jeffrey wanted a prior with $\textit{BF}[H_1 : H_2] \to 0$ (or equivalently, $\textit{BF}[H_2 : H_1] \to \infty$ ), as the information from the t-statistic grows, indicating the sample mean is as far as from the hypothesized mean and should favor $H_2$ .

To resolve the paradox when the information the t-statistic favors $H_2$ but the Bayes factor does not, Jeffreys showed that no normal prior could resolve the paradox .

But a Cauchy prior on $\mu$ , would resolve it. In this way, $\textit{BF}[H_2 : H_1]$ goes to infinity as the sample mean becomes further away from the hypothesized mean. Recall that the Cauchy prior is written as $\textsf{C}(0, r^2 \sigma^2)$ . While Jeffreys used a default of $r = 1$ , smaller values of $r$ can be used if smaller effects are expected.

The combination of the Jeffrey’s prior on $\sigma^2$ and this Cauchy prior on $\mu$ under $H_2$ is sometimes referred to as the Jeffrey-Zellener-Siow prior .

However, there is no closed form expressions for the Bayes factor under the Cauchy distribution. To obtain the Bayes factor, we must use the numerical integration or simulation methods.

We will use the function from the package to test whether the mean difference is zero in Example 5.1 (zinc), using the JZS (Jeffreys-Zellener-Siow) prior.

hypothesis testing for normal distribution

With equal prior probabilities on the two hypothesis, the Bayes factor is the posterior odds. From the output, we see this indicates that the hypothesis $H_2$ , the mean difference is different from 0, is almost 51 times more likely than the hypothesis $H_1$ that the average concentration is the same at the surface and the bottom.

To sum up, we have used the Cauchy prior as a default prior testing hypothesis about a normal mean when variances are unknown. This does require numerical integration, but it is available in the function from the package. If you expect that the effect sizes will be small, smaller values of $r$ are recommended.

It is often important to quantify the magnitude of the difference in addition to testing. The Cauchy Prior provides a default prior for both testing and inference; it avoids problems that arise with choosing a value of $n_0$ (prior sample size) in both cases. In the next section, we will illustrate using the Cauchy prior for comparing two means from independent normal samples.

5.3 Comparing Independent Means: Hypothesis Testing

In the previous section, we described Bayes factors for testing whether the mean difference of paired samples was zero. In this section, we will consider a slightly different problem – we have two independent samples, and we would like to test the hypothesis that the means are different or equal.

Example 5.2 We illustrate the testing of independent groups with data from a 2004 survey of birth records from North Carolina, which are available in the package.

The variable of interest is – the weight gain of mothers during pregnancy. We have two groups defined by the categorical variable, , with levels, younger mom and older mom.

Question of interest : Do the data provide convincing evidence of a difference between the average weight gain of older moms and the average weight gain of younger moms?

We will view the data as a random sample from two populations, older and younger moms. The two groups are modeled as:

\[\begin{equation} \begin{aligned} Y_{O,i} & \mathrel{\mathop{\sim}\limits^{\rm iid}} \textsf{N}(\mu + \alpha/2, \sigma^2) \\ Y_{Y,i} & \mathrel{\mathop{\sim}\limits^{\rm iid}} \textsf{N}(\mu - \alpha/2, \sigma^2) \end{aligned} \tag{5.2} \end{equation}\]

The model for weight gain for older moms using the subscript $O$ , and it assumes that the observations are independent and identically distributed, with a mean $\mu+\alpha/2$ and variance $\sigma^2$ .

For the younger women, the observations with the subscript $Y$ are independent and identically distributed with a mean $\mu-\alpha/2$ and variance $\sigma^2$ .

Using this representation of the means in the two groups, the difference in means simplifies to $\alpha$ – the parameter of interest.

\[(\mu + \alpha/2) - (\mu - \alpha/2) = \alpha\]

You may ask, “Why don’t we set the average weight gain of older women to $\mu+\alpha$ , and the average weight gain of younger women to $\mu$ ?” We need the parameter $\alpha$ to be present in both $Y_{O,i}$ (the group of older women) and $Y_{Y,i}$ (the group of younger women).

We have the following competing hypotheses:

$H_1: \alpha = 0 \Leftrightarrow$ The means are not different.
$H_2: \alpha \neq 0 \Leftrightarrow$ The means are different.

In this representation, $\mu$ represents the overall weight gain for all women. (Does the model in Equation (5.2) make more sense now?) To test the hypothesis, we need to specify prior distributions for $\alpha$ under $H_2$ (c.f. $\alpha = 0$ under $H_1$ ) and priors for $\mu,\sigma^2$ under both hypotheses.

Recall that the Bayes factor is the ratio of the distribution of the data under the two hypotheses.

\[\begin{aligned} \textit{BF}[H_1 : H_2] &= \frac{p(\text{data}\mid H_1)} {p(\text{data}\mid H_2)} \\ &= \frac{\iint p(\text{data}\mid \alpha = 0,\mu, \sigma^2 )p(\mu, \sigma^2 \mid H_1) \, d\mu \,d\sigma^2} {\int \iint p(\text{data}\mid \alpha, \mu, \sigma^2) p(\alpha \mid \sigma^2) p(\mu, \sigma^2 \mid H_2) \, d \mu \, d\sigma^2 \, d \alpha} \end{aligned}\]

As before, we need to average over uncertainty and the parameters to obtain the unconditional distribution of the data. Now, as in the test about a single mean, we cannot use improper or non-informative priors for $\alpha$ for testing.

Under $H_2$ , we use the Cauchy prior for $\alpha$ , or equivalently, the Cauchy prior on the standardized effect $\delta$ with the scale of $r$ :

\[\delta = \alpha/\sigma^2 \sim \textsf{C}(0, r^2)\]

Now, under both $H_1$ and $H_2$ , we use the Jeffrey’s reference prior on $\mu$ and $\sigma^2$ :

\[p(\mu, \sigma^2) \propto 1/\sigma^2\]

While this is an improper prior on $\mu$ , this does not suffer from the Bartlett’s-Lindley’s-Jeffreys’ paradox as $\mu$ is a common parameter in the model in $H_1$ and $H_2$ . This is another example of the Jeffreys-Zellner-Siow prior.

As in the single mean case, we will need numerical algorithms to obtain the Bayes factor. Now the following output illustrates testing of Bayes factors, using the Bayes inference function from the package.

We see that the Bayes factor for $H_1$ to $H_2$ is about 5.7, with positive support for $H_1$ that there is no difference in average weight gain between younger and older women. Using equal prior probabilities, the probability that there is a difference in average weight gain between the two groups is about 0.15 given the data. Based on the interpretation of Bayes factors from Table 3.5 , this is in the range of “positive” (between 3 and 20).

To recap, we have illustrated testing hypotheses about population means with two independent samples, using a Cauchy prior on the difference in the means. One assumption that we have made is that the variances are equal in both groups . The case where the variances are unequal is referred to as the Behren-Fisher problem, and this is beyond the scope for this course. In the next section, we will look at another example to put everything together with testing and discuss summarizing results.

5.4 Inference after Testing

In this section, we will work through another example for comparing two means using both hypothesis tests and interval estimates, with an informative prior. We will also illustrate how to adjust the credible interval after testing.

Example 5.3 We will use the North Carolina survey data to examine the relationship between infant birth weight and whether the mother smoked during pregnancy. The response variable, , is the birth weight of the baby in pounds. The categorical variable provides the status of the mother as a smoker or non-smoker.

We would like to answer two questions:

Is there a difference in average birth weight between the two groups?

If there is a difference, how large is the effect?

As before, we need to specify models for the data and priors. We treat the data as a random sample for the two populations, smokers and non-smokers.

The birth weights of babies born to non-smokers, designated by a subgroup $N$ , are assumed to be independent and identically distributed from a normal distribution with mean $\mu + \alpha/2$ , as in Section 5.3 .

\[Y_{N,i} \mathrel{\mathop{\sim}\limits^{\rm iid}}\textsf{Normal}(\mu + \alpha/2, \sigma^2)\]

While the birth weights of the babies born to smokers, designated by the subgroup $S$ , are also assumed to have a normal distribution, but with mean $\mu - \alpha/2$ .

\[Y_{S,i} \mathrel{\mathop{\sim}\limits^{\rm iid}}\textsf{Normal}(\mu - \alpha/2, \sigma^2)\]

The difference in the average birth weights is the parameter $\alpha$ , because

\[(\mu + \alpha/2) - (\mu - \alpha/2) = \alpha\] .

The hypotheses that we will test are $H_1: \alpha = 0$ versus $H_2: \alpha \ne 0$ .

We will still use the Jeffreys-Zellner-Siow Cauchy prior. However, since we may expect the standardized effect size to not be as strong, we will use a scale of $r = 0.5$ rather than 1.

Therefore, under $H_2$ , we have \[\delta = \alpha/\sigma \sim \textsf{C}(0, r^2), \text{ with } r = 0.5.\]

Under both $H_1$ and $H_2$ , we will use the reference priors on $\mu$ and $\sigma^2$ :

\[\begin{aligned} p(\mu) &\propto 1 \\ p(\sigma^2) &\propto 1/\sigma^2 \end{aligned}\]

The input to the base inference function is similar, but now we will specify that $r = 0.5$ .

We see that the Bayes factor is 1.44, which weakly favors there being a difference in average birth weights for babies whose mothers are smokers versus mothers who did not smoke. Converting this to a probability, we find that there is about a 60% chance of the average birth weights are different.

While looking at evidence of there being a difference is useful, Bayes factors and posterior probabilities do not convey any information about the magnitude of the effect. Reporting a credible interval or the complete posterior distribution is more relevant for quantifying the magnitude of the effect.

Using the function, we can generate samples from the posterior distribution under $H_2$ using the option.

The 2.5 and 97.5 percentiles for the difference in the means provide a 95% credible interval of 0.023 to 0.57 pounds for the difference in average birth weight. The MCMC output shows not only summaries about the difference in the mean $\alpha$ , but the other parameters in the model.

In particular, the Cauchy prior arises by placing a gamma prior on $n_0$ and the conjugate normal prior. This provides quantiles about $n_0$ after updating with the current data.

The row labeled effect size is the standardized effect size $\delta$ , indicating that the effects are indeed small relative to the noise in the data.

Figure 5.4: Estimates of effect under H2

Figure 5.4 shows the posterior density for the difference in means, with the 95% credible interval indicated by the shaded area. Under $H_2$ , there is a 95% chance that the average birth weight of babies born to non-smokers is 0.023 to 0.57 pounds higher than that of babies born to smokers.

The previous statement assumes that $H_2$ is true and is a conditional probability statement. In mathematical terms, the statement is equivalent to

\[P(0.023 < \alpha < 0.57 \mid \text{data}, H_2) = 0.95\]

However, we still have quite a bit of uncertainty based on the current data, because given the data, the probability of $H_2$ being true is 0.59.

\[P(H_2 \mid \text{data}) = 0.59\]

Using the law of total probability, we can compute the probability that $\mu$ is between 0.023 and 0.57 as below:

\[\begin{aligned} & P(0.023 < \alpha < 0.57 \mid \text{data}) \\ = & P(0.023 < \alpha < 0.57 \mid \text{data}, H_1)P(H_1 \mid \text{data}) + P(0.023 < \alpha < 0.57 \mid \text{data}, H_2)P(H_2 \mid \text{data}) \\ = & I( 0 \text{ in CI }) P(H_1 \mid \text{data}) + 0.95 \times P(H_2 \mid \text{data}) \\ = & 0 \times 0.41 + 0.95 \times 0.59 = 0.5605 \end{aligned}\]

Finally, we get that the probability that $\alpha$ is in the interval, given the data, averaging over both hypotheses, is roughly 0.56. The unconditional statement is the average birth weight of babies born to nonsmokers is 0.023 to 0.57 pounds higher than that of babies born to smokers with probability 0.56. This adjustment addresses the posterior uncertainty and how likely $H_2$ is.

To recap, we have illustrated testing, followed by reporting credible intervals, and using a Cauchy prior distribution that assumed smaller standardized effects. After testing, it is common to report credible intervals conditional on $H_2$ . We also have shown how to adjust the probability of the interval to reflect our posterior uncertainty about $H_2$ . In the next chapter, we will turn to regression models to incorporate continuous explanatory variables.

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4.1 The normal distribution

Here, you will look at the concept of normal distribution and the bell-shaped curve. The peak point (the top of the bell) represents the most probable occurrences, while other possible occurrences are distributed symmetrically around the peak point, creating a downward-sloping curve on either side of the peak point.

The cartoon shows a bell-shaped curve. The x-axis is titled ‘How high the hill is’ and the y-axis is titled ‘Number of hills’. The top of the bell-shaped curve is labelled ‘Average hill’, but on the lower right tail of the bell-shaped curve is labelled ‘Big hill’.

In order to test hypotheses, you need to calculate the test statistic and compare it with the value in the bell curve. This will be done by using the concept of ‘normal distribution’.

A normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more likely to occur than data far from it. In graph form, a normal distribution appears as a bell curve. The values in the x-axis of the normal distribution graph represent the z-scores. The test statistic that you wish to use to test the set of hypotheses is the z-score . A z-score is used to measure how far the observation (sample mean) is from the 0 value of the bell curve (population mean). In statistics, this distance is measured by standard deviation. Therefore, when the z-score is equal to 2, the observation is 2 standard deviations away from the value 0 in the normal distribution curve.

A symmetrical graph reminiscent of a bell showing normal distribution.

A symmetrical graph reminiscent of a bell. The top of the bell-shaped curve appears where the x-axis is at 0. This is labelled as Normal distribution.

Hypothesis Testing with the Normal Distribution

Contents Toggle Main Menu 1 Introduction 2 Test for Population Mean 3 Worked Example 3.1 Video Example 4 Approximation to the Binomial Distribution 5 Worked Example 6 Comparing Two Means 7 Workbooks 8 See Also

Introduction

When constructing a confidence interval with the standard normal distribution, these are the most important values that will be needed.

Distribution of Sample Means

where $\mu$ is the true mean and $\mu_0$ is the current accepted population mean. Draw samples of size $n$ from the population. When $n$ is large enough and the null hypothesis is true the sample means often follow a normal distribution with mean $\mu_0$ and standard deviation $\frac{\sigma}{\sqrt{n}}$. This is called the distribution of sample means and can be denoted by $\bar{X} \sim \mathrm{N}\left(\mu_0, \frac{\sigma}{\sqrt{n}}\right)$. This follows from the central limit theorem .

The $z$-score will this time be obtained with the formula \[Z = \dfrac{\bar{X} - \mu_0}{\frac{\sigma}{\sqrt{n}}}.\]

So if $\mu = \mu_0, X \sim \mathrm{N}\left(\mu_0, \frac{\sigma}{\sqrt{n}}\right)$ and $ Z \sim \mathrm{N}(0,1)$.

The alternative hypothesis will then take one of the following forms: depending on what we are testing.

Worked Example

An automobile company is looking for fuel additives that might increase gas mileage. Without additives, their cars are known to average $25$ mpg (miles per gallons) with a standard deviation of $2.4$ mpg on a road trip from London to Edinburgh. The company now asks whether a particular new additive increases this value. In a study, thirty cars are sent on a road trip from London to Edinburgh. Suppose it turns out that the thirty cars averaged $\overline{x}=25.5$ mpg with the additive. Can we conclude from this result that the additive is effective?

We are asked to show if the new additive increases the mean miles per gallon. The current mean $\mu = 25$ so the null hypothesis will be that nothing changes. The alternative hypothesis will be that $\mu > 25$ because this is what we have been asked to test.

\begin{align} &H_0:\mu=25. \\ &H_1:\mu>25. \end{align}

Now we need to calculate the test statistic. We start with the assumption the normal distribution is still valid. This is because the null hypothesis states there is no change in $\mu$. Thus, as the value $\sigma=2.4$ mpg is known, we perform a hypothesis test with the standard normal distribution. So the test statistic will be a $z$ score. We compute the $z$ score using the formula \[z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n} } }.\] So \begin{align} z&=\frac{\overline{x}-25}{\frac{2.4}{\sqrt{30} } }\\ &=1.14 \end{align}

We are using a $5$% significance level and a (right-sided) one-tailed test, so $\alpha=0.05$ so from the tables we obtain $z_{1-\alpha} = 1.645$ is our test statistic.

As $1.14<1.645$, the test statistic is not in the critical region so we cannot reject $H_0$. Thus, the observed sample mean $\overline{x}=25.5$ is consistent with the hypothesis $H_0:\mu=25$ on a $5$% significance level.

Video Example

In this video, Dr Lee Fawcett explains how to conduct a hypothesis test for the mean of a single distribution whose variance is known, using a one-sample z-test.

Approximation to the Binomial Distribution

A supermarket has come under scrutiny after a number of complaints that its carrier bags fall apart when the load they carry is $5$kg. Out of a random sample of $200$ bags, $185$ do not tear when carrying a load of $5$kg. Can the supermarket claim at a $5$% significance level that more that $90$% of the bags will not fall apart?

Let $X$ represent the number of carrier bags which can hold a load of $5$kg. Then $X \sim \mathrm{Bin}(200,p)$ and \begin{align}H_0&: p = 0.9 \\ H_1&: p > 0.9 \end{align}

We need to calculate the mean $\mu$ and variance $\sigma ^2$.

\[\mu = np = 200 \times 0.9 = 180\text{.}\] \[\sigma ^2= np(1-p) = 18\text{.}\]

Using the normal approximation to the binomial distribution we obtain $Y \sim \mathrm{N}(180, 18)$.

\[\mathrm{P}[X \geq 185] = \mathrm{P}\left[Z \geq \dfrac{184.5 - 180}{4.2426} \right] = \mathrm{P}\left[Z \geq 1.0607\right] \text{.}\]

Because we are using a one-tailed test at a $5$% significance level, we obtain the critical value $Z=1.645$. Now $1.0607 < 1.645$ so we cannot accept the alternative hypothesis. It is not true that over $90$% of the supermarket's carrier bags are capable of withstanding a load of $5$kg.

Comparing Two Means

When we test hypotheses with two means, we will look at the difference $\mu_1 - \mu_2$. The null hypothesis will be of the form

where $a$ is a constant. Often $a=0$ is used to test if the two means are the same. Given two continuous random variables $X_1$ and $X_2$ with means $\mu_1$ and $\mu_2$ and variances $\frac{\sigma_1^2}{n_1}$ and $\frac{\sigma_2^2}{n_2}$ respectively \[\mathrm{E} [\bar{X_1} - \bar{X_2} ] = \mathrm{E} [\bar{X_1}] - \mathrm{E} [\bar{X_2}] = \mu_1 - \mu_2\] and \[\mathrm{Var}[\bar{X_1} - \bar{X_2}] = \mathrm{Var}[\bar{X_1}] - \mathrm{Var}[\bar{X_2}]=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}\text{.}\] Note this last result, the difference of the variances is calculated by summing the variances.

We then obtain the $z$-score using the formula \[Z = \frac{(\bar{X_1}-\bar{X_2})-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\text{.}\]

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.

Tests concerning a single sample
Tests concerning two samples

Selecting a Hypothesis Test

Hypothesis Testing

First Online: 26 April 2022

Cite this chapter

Melissa Whatley ORCID: orcid.org/0000-0002-7073-6772 2

Part of the book series: Springer Texts in Education ((SPTE))

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Drawing from the properties of the normal distribution, this chapter introduces key concepts in hypothesis testing, including the Central Limit Theorem, the sampling distribution, the expected value of the mean, and the estimated standard error. These concepts are illustrated through three types of t-tests: one-sample t-test, two-samples t-tests, and dependent samples t-tests.

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The full details of the Central Limit Theorem are beyond the scope of this book, but I encourage you to explore them further in a more advanced statistics textbook, such as those listed in the recommended reading list at the end of this chapter.

Note here that we also assume that our sample is valid.

When I teach Type 1 and Type 2 error, I often draw a parallel to my dogs, Taca, Bernice, Lola, and Fritsi, barking at the front door of my house. Sometimes, they bark at the front door even when no one is there (perhaps there are ghosts in my neighborhood?). This is a Type 1 error. They detect an effect (effect = someone at the door) that does not exist in reality. Other times, they fail to bark at the front door even when there is someone there (a couple weeks ago, I ordered pizza delivery for dinner, and they did not even notice when the delivery person arrived to drop the pizza off). This is a Type 2 error. They do not detect an effect even though it exists in reality.

Technically, this formula is only appropriate when you do not know the population standard deviation and have to estimate the standard error using the sample standard deviation, an assumption we make throughout this chapter.

These are not the only additional mean-comparison hypothesis tests, but they are likely the most useful for readers of this book.

The formula for the estimated standard error of the difference between two independent sample means when sample sizes are not equal is considerably more complex:

$s_{\overline{x}1 - \overline{x}2} = \sqrt {\frac{SS_1 + SS_2 }{{n_1 + n_2 - 2}}\left( {\frac{1}{n_1 } + \frac{1}{n_2 }} \right)}$

This formula also assumes that population standard deviations are equal. Here, $SS_1$ and $SS_2$ refer to the sum of squared deviations for each sample, while $n_1$ and $n_2$ are the two sample sizes. Essentially, what this formula does is weight the standard error to account for the fact that the sample sizes are unequal. Differences in sample sizes are especially problematic when one sample’s standard error is dramatically different from the standard error of the other sample (a violation of the homogeneity of variance assumption). This can lead to very misleading results. When you calculate a two-samples t- test using a statistical software program, this adjustment for sample size, if needed, is made for you automatically. It also is possible to compute a standard error that allows for unequal variances between samples.

Additional Examples

Cartwright, C., Stevens, M., & Schneider, K. (2021). Constructing the learning outcomes with intercultural assessment: A 3-year study of a graduate study abroad and glocal experience programs. Frontiers: The Interdisciplinary Journal of Study Abroad , 33 (1), 82–105.

Echcharfy, M. (2020). Intercultural learning in Moroccan higher education: A comparison between teachers’ perceptions and students’ expectations. International Journal of ResEarch in English Education, 5 (1), 19–35.

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Yang, L., Borrowman, L., Tan, M. Y., & New, J. Y. (2020). Expectations in transition: Students’ and teachers’ expectations of university in an international branch campus. Journal of Studies in International Education, 24 (3), 352–370.

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How to Do Hypothesis Testing with Normal Distribution

Hypothesis tests compare a result against something you already believe is true. Let X 1 , X 2 , … ⁡ , X n be n independent random variables with equal expected value μ and standard deviation σ . Let X be the mean of these n random variables, so

The stochastic variable X has an expected value μ and a standard deviation σ n . You want to perform a hypothesis test on this expected value. You have a null hypothesis H 0 : μ = μ 0 and three possible alternative hypotheses: H a : μ < μ 0 , H a : μ > μ 0 or H a : μ ≠ μ 0 . The first two alternative hypotheses belong to what you call a one-sided test, while the latter is two-sided.

In hypothesis testing, you calculate using the alternative hypothesis in order to say something about the null hypothesis.

Hypothesis Testing (Normal Distribution)

Note! For two-sided testing, multiply the p -value by 2 before checking against the critical region.

As the production manager at the new soft drink factory, you are worried that the machines don’t fill the bottles to their proper capacity. Each bottle should be filled with 0 . 5 L soda, but random samples show that 48 soda bottles have an average of 0 . 4 8 L , with an empirical standard deviation of 0 . 1 . You are wondering if you need to recalibrate the machines so that they become more precise.

This is a classic case of hypothesis testing by normal distribution. You now follow the instructions above and select 1 0 % level of significance, since it is only a quantity of soda and not a case of life and death.

The alternative hypothesis in this case is that the bottles do not contain 0 . 5 L and that the machines are not precise enough. This thus becomes a two-sided hypothesis test and you must therefore remember to multiply the p -value by 2 before deciding whether the p -value is in the critical region. This is because the normal distribution is symmetric, so P ( X ≥ k ) = P ( X ≤ − k ) . Thus it is just as likely to observe an equally extremely high value as an equally extreme low:

so H 0 must be kept, and the machines are deemed to be fine as is.

Had the p -value been less than the level of significance, that would have meant that the calibration represented by the alternative hypothesis would be significantly better for the business.

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Hypothesis tests for the normal distribution can be conducted in a very similar way to binomial distribution , e xcept this time we switch our test statistic. These tests are useful as again they help us test claims of normally distributed items.

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How do we carry out a hypothesis test for normal distribution?

When we hypothesis test for the mean of a normal distribution we think about looking at the mean of a sample from a population.

So for a random sample of size n of a population, taken from the random variable $X \sim N(\mu, \sigma^2)$ , the sample mean $\bar{X}$ can be normally distributed by $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$ .

Let's look at an example.

The weight of crisps is each packet is normally distributed with a standard deviation of 2.5g.

The crisp company claims that the crisp packets have a mean weight of 28g. There were numerous complaints that each crisp packet weighs less than this. Therefore a trading inspector investigated this and found in a sample of 50 crisp packets, the mean weight was 27.2g.

Using a 5% significance level and stating the hypothesis, clearly test whether or not the evidence upholds the complaints.

This is an example of a one tailed test. Let's look at an example of a two tailed test.

A machine produces circular discs with a radius R, where R is normally distributed with a mean of 2cm and a standard deviation of 0.3cm.

The machine is serviced and after the service, a random sample of 40 discs is taken to see if the mean has changed from 2cm. The radius is still normally distributed with a standard deviation of 0.3 cm.

The mean is found to be 1.9cm.

Has the mean changed? Test this to a 5% significance level.

Step 5 may be confusing – do we carry out the calculation with $P(\bar{X} \leq \bar{x})$ or $P(\bar{X} \geq \bar{x})$? As a general rule of thumb if the value is between 0 and the mean, then we use $P(\bar{X} \leq \bar{x})$ . If it is greater than the mean then we use $P(\bar{X} \geq \bar{x})$ .

How about finding critical values and critical regions?

This is the same idea as in binomial distribution . However, in normal distribution, a calculator can make our lives easier.

The distributions menu has an option called inverse normal.

Here, we enter the significance level (Area), the mean ($\mu$ ) and the standard deviation ($\sigma$ ).

The calculator will give us an answer. Let's have a look at an example below.

Wheels are made to measure for a bike. The diameter of the wheel is normally distributed with a mean of 40cm and a standard deviation of 5cm. Some people think that their wheels are too small. Find the critical value of this to a 5% significance level.

In our calculator, in the inverse normal function, we need to enter:

If we perform the inverse normal function we get 31.775732 .

So that is our critical value and our critical region is $X \leq 31.775732$ .

Let's look at an example with two tails.

Normal distribution hypothesis test two tailed test studysmarter

Hypothesis Test for Normal Distribution - Key takeaways

When we hypothesis test for a normal distribution we are trying to see if the mean is different from the mean stated in the null hypothesis.
We use the sample mean which is $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$ .
In two-tailed tests we divide the significance level by two and test on both tails.
When finding critical values we use the calculator inverse normal function entering the area as the significance level.
For two-tailed tests we need to find two critical values on either end of the distribution.

Frequently Asked Questions about Normal Distribution Hypothesis Test

--> how do you test a hypothesis for a normal distribution, --> is hypothesis testing only for a normal distribution.

No, pretty much any distribution can be used when testing a hypothesis. The two distributions that you learn at A-Level are Normal and Binomial.

--> What statistical hypothesis can be tested in the means of a normal distribution?

We test whether or not the data can support the value of a mean being too low or too high.

What is our test statistic with normal distribution?

What calculator tool do we use to work backwards with normal distribution?

The Inverse Normal

A coach thinks his athletes will achieve less than 12 seconds in their 100 metre race. His assistant thinks they won't be this fast. If this claim was tested is this a one or two-tailed test?

One-tailed test

How do we find the critical region of a normal distribution?

By using the calculator inverse normal setting.

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Available Hypothesis Tests

In addition to the previous functions, Statistics and Machine Learning Toolbox™ functions are available for analysis of variance (ANOVA), which perform hypothesis tests in the context of linear modeling.

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Hypothesis tests about the variance

by Marco Taboga , PhD

This page explains how to perform hypothesis tests about the variance of a normal distribution, called Chi-square tests.

We analyze two different situations:

when the mean of the distribution is known;

when it is unknown.

Depending on the situation, the Chi-square statistic used in the test has a different distribution.

At the end of the page, we propose some solved exercises.

Table of contents

Normal distribution with known mean

The null hypothesis, the test statistic, the critical region, the decision, the power function, the size of the test, how to choose the critical value, normal distribution with unknown mean, solved exercises.

The assumptions are the same previously made in the lecture on confidence intervals for the variance .

The sample is drawn from a normal distribution .

A test of hypothesis based on it is called a Chi-square test .

Otherwise the null is not rejected.

We explain how to do this in the page on critical values .

We now relax the assumption that the mean of the distribution is known.

See the comments on the choice of the critical value made for the case of known mean.

Below you can find some exercises with explained solutions.

Suppose that we observe 40 independent realizations of a normal random variable.

we run a Chi-square test of the null hypothesis that the variance is equal to 1;

Make the same assumptions of Exercise 1 above.

If the unadjusted sample variance is equal to 0.9, is the null hypothesis rejected?

How to cite

Please cite as:

Taboga, Marco (2021). "Hypothesis tests about the variance", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/hypothesis-testing-variance.

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COMMENTS

9.4: Distribution Needed for Hypothesis Testing
Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's $t$-distribution. (Remember, use a Student's $t$-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.)
Normal Distribution Hypothesis Tests
When to do a Normal Hypothesis Test. There are two types of hypothesis tests you need to know about: binomial distribution hypothesis tests and normal distribution hypothesis tests.In binomial hypothesis tests, you are testing the probability parameter p.In normal hypothesis tests, you are testing the mean parameter \mu.This gives us a key difference that we can use to determine what test to ...
5.3.2 Normal Hypothesis Testing
What steps should I follow when carrying out a hypothesis test for the mean of a normal distribution? Following these steps will help when carrying out a hypothesis test for the mean of a normal distribution: Step 1. Define the distribution of the population mean usually Step 2. Write the null and alternative hypotheses clearly using the form
9.3 Probability Distribution Needed for Hypothesis Testing
Assumptions. When you perform a hypothesis test of a single population mean μ using a normal distribution (often called a z-test), you take a simple random sample from the population. The population you are testing is normally distributed, or your sample size is sufficiently large.You know the value of the population standard deviation, which, in reality, is rarely known.
7.4.1
Step 1: Check assumptions and write hypotheses. The assumption here is that the sampling distribution is approximately normal. From the given StatKey output, the randomization distribution is approximately normal. \ (H_0\colon p=0.50\) \ (H_a\colon p>0.50\) 2. Calculate the test statistic.
9.3 Distribution Needed for Hypothesis Testing
Earlier in the course, we discussed sampling distributions. Particular distributions are associated with hypothesis testing. Perform tests of a population mean using a normal distribution or a Student's t-distribution. (Remember, use a Student's t-distribution when the population standard deviation is unknown and the distribution of the sample mean is approximately normal.)
Hypothesis Testing
One sample z-test . Assume data are independently sampled from a normal distribution with unknown mean μ and known variance σ 2 = 9. Make an initial assumption that μ = 65. Specify the hypothesis: H 0: μ = 65 H A: μ ≠ 65. z-statistic: 3.58. z-statistic follow N(0,1) distribution
Distribution Needed for Hypothesis Testing
We perform tests of a population proportion using a normal distribution (usually n is large or the sample size is large). If you are testing a single population mean, the distribution for the test is for means: ¯ X~N(μX , σX √n) or tdf. The population parameter is μ. The estimated value (point estimate) for μ is ¯ x, the sample mean.
Chapter 5 Hypothesis Testing with Normal Populations
5.1 Bayes Factors for Testing a Normal Mean: variance known. Now we show how to obtain Bayes factors for testing hypothesis about a normal mean, where the variance is known.To start, let's consider a random sample of observations from a normal population with mean $\mu$ and pre-specified variance $\sigma^2$.We consider testing whether the population mean $\mu$ is equal to $m_0$ or not.
Data analysis: hypothesis testing: 4.1 The normal distribution
A normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more likely to occur than data far from it. In graph form, a normal distribution appears as a bell curve. The values in the x-axis of the normal distribution graph represent the z-scores. The test statistic that you wish ...
PDF §5.1 HYPOTHESIS TESTS USING NORMAL DISTRIBUTIONS
Definition (Normal Distribution) The normal distributions are a family of distribution curves. Each member of the family is specified by two parameters: the standard deviation, denoted by σ. normal distribution follows a bell-shaped curve. For shorthand we often use the notation N (μ, σ) to specify a normal distribution with parameters μ ...
Numeracy, Maths and Statistics
The alternative hypothesis will be that $\mu > 25$ because this is what we have been asked to test. \begin{align} &H_0:\mu=25. \\ &H_1:\mu>25. \end{align} Now we need to calculate the test statistic. We start with the assumption the normal distribution is still valid. This is because the null hypothesis states there is no change in $\mu$.
Introduction to Hypothesis testing for Normal distribution
Introduction to Hypothesis testing for Normal distributionIn this tutorial, we learn how to conduct a hypothesis test for normal distribution using p values ...
Normal Distribution
Hypothesis testing. Hypothesis testing guide; Null vs. alternative hypotheses; Statistical significance; p value; Type I & Type II errors; Statistical power; ... The normal distribution is a probability distribution, so the total area under the curve is always 1 or 100%. The formula for the normal probability density function looks fairly ...
Hypothesis Testing
These properties are key to hypothesis testing and, for that matter, most of the statistics that you will see in this book. It is worth noting that a standardized version of the normal distribution, called the z-distribution, rescales continuous variables so that the mean is set to 0 and each standard deviation is equal to 1. This is common ...
How to Do Hypothesis Testing with Normal Distribution
Hypothesis Testing (Normal Distribution) 1. You set up the null hypothesis H 0 against an alternative hypothesis H A. H 0: μ = μ 0 against H a: μ > μ 0 (possibly H a: μ < μ 0, H a: μ ≠ μ 0 ). 2. Then, you do an experiment and find that the average value is x. Next, you calculate the probability P ( X ≥ x) for the alternative ...
Normal Distribution Hypothesis Test: Explanation & Example
When we hypothesis test for a normal distribution we are trying to see if the mean is different from the mean stated in the null hypothesis. We use the sample mean which is $\bar{X} \sim N(\mu, \frac{\sigma^2}{n})$. In two-tailed tests we divide the significance level by two and test on both tails.
Available Hypothesis Tests
Tests if a sample comes from a normal distribution with known variance and specified mean, against the alternative that it does not have that mean. Note In addition to the previous functions, Statistics and Machine Learning Toolbox™ functions are available for analysis of variance (ANOVA), which perform hypothesis tests in the context of ...
Hypothesis tests about the variance
This page explains how to perform hypothesis tests about the variance of a normal distribution, called Chi-square tests. We analyze two different situations: when it is unknown. Depending on the situation, the Chi-square statistic used in the test has a different distribution. At the end of the page, we propose some solved exercises.
Tackling Non-Normal Data in Hypothesis Tests
Permutation tests offer a data-driven approach to hypothesis testing. By shuffling the data labels and recalculating the test statistic for each permutation, you can create a distribution of the ...

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Module 9: Hypothesis Testing With One Sample

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An Introduction to Bayesian Thinking

5.1 Bayes Factors for Testing a Normal Mean: variance known

5.2 Comparing Two Paired Means using Bayes Factors

5.3 Comparing Independent Means: Hypothesis Testing

5.4 Inference after Testing

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4.1 The normal distribution

Hypothesis Testing with the Normal Distribution

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How to Do Hypothesis Testing with Normal Distribution

Hypothesis Testing (Normal Distribution)

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How do we carry out a hypothesis test for normal distribution?

How about finding critical values and critical regions?

Hypothesis Test for Normal Distribution - Key takeaways

Frequently Asked Questions about Normal Distribution Hypothesis Test

--> What statistical hypothesis can be tested in the means of a normal distribution?

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