chapter 8 hypothesis testing answers

Introduction to Statistics

Chapter 8 hypothesis testing.

Learning Outcome

Perform hypotheses testing involving one population mean, one population proportion, and one population standard deviations/variance.

This chapter introduces the statistical method of hypothesis testing to test a given claim about a population parameter, such as proportion, mean, standard deviation, or variance. This method combines the concepts covered in the previous chapters, including sampling distribution, standard error, critical scores, and probability theory.

8.1 Hypothesis Testing

In statistics, a hypothesis is a claim or statement about a property of a population.

A hypothesis test (or test of significance) is a procedure for testing a claim about a property of a population.

The null hypothesis ( \(H_0\) ) is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value.

The alternative hypothesis ( \(H_A\) ) is a statement that the parameter has a value that somehow differs from the null hypothesis.

Purpose of a Hypothesis Test

The purpose of a hypothesis test is to determine how plausible the null hypothesis is. At the start of a hypothesis test, we assume that the null hypothesis is true. Then we look at the evidence, which comes from data that have been collected. If the data strongly indicate that the null hypothesis is false, we abandon our assumption that it is true and believe the alternate hypothesis instead. This is referred to as rejecting the null hypothesis.

The evidence comes in the form of a test statistic. When the difference between the test statistic (such as, \(z\) or \(t\) scores) and the value in the null hypothesis is sufficiently large, we reject the null hypothesis.

\[ \fbox{Assume the null hypothesis is true} \xrightarrow{} \fbox{consider the evidence} \xrightarrow{} \fbox{decide whether to accept or reject the null hypothesis} \]

Probability of getting head from a single toss of coin, \(p = 0.5\) .

Therefore, expected value of the number of heads from \(20\) tosses = \(10\) .

Suppose, on your first trial, you have tossed a coin \(20\) times and seen \(15\) heads, \(\hat p = 0.75\) . On your second trial, you have tossed a coin \(20\) times again and seen \(12\) heads, \(\hat p = 0.60\) .

Is the coin fair, or is it biased towards heads?

Null and Alternative Hypotheses

Null hypothesis \((H_0)\) : states that any deviation from what was expected is due to chance error (i.e. the coin is fair).

Alternative hypothesis \((H_A)\) : asserts that the observed deviation is too large to be explained by chance alone (i.e. the coin is biased towards heads).

\[ H_0: p = 0.5 \\ H_A: p > 0.5 \]

Now, what is the probability of \(p \ge 0.75?\) What is the probability of \(p \ge 0.60?\)

From normal approximation of the sampling distribution of \(\hat p\) ,

\[ \begin{align} p &= 0.5 \\ se &= \sqrt{(0.5)(0.5)/20} = 0.112 \\ z_1 &= (0.75 - 0.50)/0.112 = 2.236 \\ \\ z_2 &= (0.60 - 0.50)/0.112 = 0.893 \\ \\ \text {p-value} &= \begin{cases} P(z\ge 2.236) &= 0.0127 \\ P(z\ge 0.893) &= 0.1860 \end{cases} \end{align} \]

We see that the farther the test statistic is from the values specified by \(H_0\) , the less likely the difference is due to chance - and the less plausible becomes. The question then is: How big should the difference be before we reject \(H_0\) ?

To answer this question, we need methods that enable us to calculate just how plausible \(H_0\) is. Hypothesis tests provide these methods taking into account things such as the size of the sample and the amount of spread in the distribution.

Interpretation of \(\text{p-value}\)

A \(\text{p-value}\) is the probability of obtaining the observed effect (or larger) under a “null hypothesis”. Thus, a \(\text{p-value}\) that is very small indicates that the observed effect is very unlikely to have arisen purely by chance, and therefore provides evidence against the null hypothesis.

It has been common practice to interpret a \(\text{p-value}\) by examining whether it is smaller than particular threshold values or “significance level”. In particular, \(\text{p-values}\) less than \(5\%\) are often reported as “statistically significant”, and interpreted as being small enough to justify rejection of the null hypothesis. By definition, the significance level \(\alpha\) is the probability of mistakenly rejecting the null hypothesis when it is true.

\[\textbf {Significance level } \alpha = P \textbf { (rejecting } H_0 \textbf { when } H_0 \textbf { is true)} \]

In common practice, \(\alpha\) is set at \(10\%, 5\%\) or \(1\%\) .

In the coin toss example:

p-value = \(1.27\%\) which is less than the \(5\%\) significance level.

Therefore, the result is statistically significant.

Conclusion: The coin is biased towards heads.

Type I and Type II Errors

When testing a null hypothesis, sometimes the test comes to a wrong conclusion by rejecting it or failing to reject it. There are two kinds of errors: type I and type II errors.

• \(\textbf {Type I error} :\) The error of rejecting the null hypothesis when it is actually true.

\(\alpha = P (\textbf{type I error}) = P (\text{rejecting } H_0 \text{ when } H_0 \text{ is true } )\)

The probability of \(\text {Type I}\) can be minimized by choosing a smaller \(\alpha\) .

• \(\textbf {Type II error} :\) The error of failing to reject the null hypothesis when it is actually false.

\(\beta = P (\textbf{type II error}) = P (\text{failing to reject } H_0 \text{ when } H_0 \text{ is false} )\)

The probability of \(\text {Type II}\) can be minimized by choosing a larger sample size \(n\) .

In general, a \(\text{Type I}\) error is more serious than a \(\text{Type II}\) error. This is because a \(\text{Type I}\) error results in a false conclusion, while a \(\text{Type II}\) error results only in no conclusion. Ideally, we would like to minimize the probability of both errors. Unfortunately, with a fixed sample size, decreasing the probability of one type increases the probability of the other.

Hypothesis tests are often designed so that the probability of a \(\text{Type I}\) error will be acceptably small, often \(0.05\) or \(0.01\) . This value is called the significance level of the test.

Significance Level

Notice that if \(H_0\) is actually true, but \(\hat p\) falls in the critical region, then a \(\text{Type I}\) Error occurs. We begin a hypothesis test by setting the probability of a \(\text{Type I}\) Error. This value is called the significance level and is denoted by \(\alpha\) . The area of critical region is equal to \(\alpha\) . The choice of \(\alpha\) is determined by how strong we require the evidence against \(H_0\) to be in order to reject it. The smaller the value of \(\alpha\) , the stronger we require the evidence to be.

Critical Value Method

In a hypothesis test, the critical value(s) separates the critical region (where we reject the null hypothesis) from the values of the test statistic that do not lead to rejection of the null hypothesis.

With the critical value method of testing hypothesis, we make a decision by comparing the test statistic to the critical value(s).

Stating the Conclusion in a Hypothesis Test

If the null hypothesis is rejected, the conclusion of the hypothesis test is straightforward: We conclude that the alternate hypothesis, \(H_0\) , is true.

If the null hypothesis is not rejected, we say that there is not enough evidence to conclude that the alternate hypothesis, \(H_a\) , is true. This is not saying the null hypothesis is true. What we are saying is that the null hypothesis might be true.

One-sided and two-sided tests

If the researchers are only interested in showing an increase or a decrease, but not both, use a one-sided test . If the researchers would be interested in any difference from the null value - an increase or decrease - then the test should be two-sided .

After observing data, it is tempting to turn a two-sided test into a one-sided test. Hypotheses must be set up before observing the data. If they are not, the test must be two-sided.

Steps of a Formal Test of Hypothesis

Follow these seven steps when carrying out a hypothesis test.

• State the name of the test being used.
• Verify conditions to ensure the standard error estimate is reasonable and the point estimate follows the appropriate distribution and is unbiased.
• Write the hypotheses and set them up in mathematical notation.
• Identify the significance level \(\alpha\) .
• Calculate the test statistics (e.g.  \(z\) ), using an appropriate point estimate of the paramater of interest and its standard error. \[\text{test statistics} = \frac{\text{point estimate - null value}}{\text{SE of estimate}}\]
• Find the \(\text{p-value}\) , compare it to \(\alpha\) , and state whether to reject or not reject the null hypothesis.
• Write your conclusion in context.

8.2 Power of a Hypothesis Test (Optional)

The power of a hypothesis test is the probability \(1-\beta\) of rejecting a false null hypothesis. The value of the power is computed by using a particular significance level \(\alpha\) and a particular value of the population parameter that is an alternative to the value of assumed true in the null hypothesis.

In practice, statistical studies are commonly designed with a statistical power of at least \(80%\) .

Post-hoc Power Calculation for One Study Group vs. Population

\[ H_0 : p = P_0 \\ H_A : p \ne P_0 \\ \] \[ \begin{align} P_0 &= \text{proportion of population} \\ P_1 &= \text{proportion observed from the data (an alternative population)} \\ N &= \text{sample size} \\ \alpha &= \text{probability of type I error} \\ \beta &= \text{probability of type II error} \\ z &= \text{critical z score for a given } \alpha \text { or } \beta \end{align} \]

Suppose, \(P_1\) is an alternative to the value assumed in \(H_0\) .

Under \(H_0\) ,

\[ P'_0 = P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \]

Under \(H_A\) ,

\[ \therefore z_{\beta} = \dfrac{P'_0 - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} = \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \\ \\ P(\textbf{Type II error}) = \beta = \Phi \left \{ \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \right \} \\ \]

\[ \begin{align} \text{where,} \\ Q_0 &= 1 - P_0 \\ Q_1 &= 1 - P_1 \\ \Phi &= \text{cumulative normal distribution function} \end{align} \]

Example: Calculate statistical power for various alternative hypotheses.

\[ Suppose, \begin{cases} H_0: p = 0.5 \\ H_A: p \ne 0.5 \\ P(\text{type I error}) = \alpha = 0.05 \\ \text{Critical z score, } z_{1 - \alpha/2} = 1.96 \\ N = 14 \\ \end{cases} \]

\[ \begin{array}{r|r|r} P_1 & \Phi(z_{\beta}) = \beta & 1-\beta \\ \hline 0.6 & \Phi(1.2367) = 0.8919 & 0.1081 \\ 0.7 & \Phi(0.5055) = 0.6934 & 0.3066 \\ 0.8 & \Phi(-0.3562) = 0.3608 & 0.6392 \\ 0.9 & \Phi(-1.7222) = 0.0425 & 0.9575 \\ \hline \end{array} \]

Example: Sample size calculation to achieve power (when \(P_0\) and \(P_1\) are known)

\[ \begin{align} z_{\beta} = \Phi^{-1}(\beta) &= \dfrac{\Bigg( P_0 + z_{1-\alpha/2} \cdot \sqrt{\dfrac{P_0Q_0}{N}} \Bigg) - P_1}{\sqrt{\dfrac{P_1Q_1}{N}}} \\ \\ z_{1-\alpha/2} &= 1.96 \\ 1- \beta &= 0.8 \\ \\ \Phi^{-1}(0.2) = -0.84 &= \dfrac{\Bigg( 0.5 + 1.96 \cdot \sqrt{\dfrac{(0.5) (0.5)}{N}} \Bigg) - 0.9}{\sqrt{\dfrac{(0.9)(0.1)}{N}}} \\ \implies N &= \Bigg( \dfrac{1.96\sqrt{0.25} + 0.84\sqrt{0.09}}{0.4} \Bigg)^2 \\ &\approx 10 \end{align} \]

Example: Sample size calculation to achieve power (when \(P_0\) and \(P_1\) are unknown)

\[ N = \Bigg( \dfrac{z_{1-\alpha/2} + z_{1-\beta}}{ES} \Bigg)^2 \\ \] where,

\[ \text{effect size, } ES = \dfrac{|P_1-P_0|}{\sqrt{P_0Q_0}} \]

Statistical power and design of experiment: When designing an experiment, it is essential to determine the minimize sample size that would be needed to detect an acceptable difference between the true value of the population parameter and what is observed from the data. A \(5\%\) significance level \((\alpha)\) and a statistical power of at least \(80\%\) are common requirements for determining that a hypothesis test is effective.

8.3 Inference for a Single Proportion

Conduct a formal hypothesis test of a claim about a population proportion \(p\) .

Requirements:

• The sample observations are simple random sample.
• The trials are independent with two possible outcomes.
• The sampling distribution for \(\hat p\) , taken from a sample of size \(n\) from a population with a true proportion \(p\) , is nearly normal when the sample observations are independent and we expect to see at least \(10\) successes and \(10\) failures in our sample, i.e.  \(np \ge 10\) and \(n(1-p) \ge 10\) . This is called the success-failure condition. If the conditions are met, then the sampling distribution of \(\hat p\) is nearly normal with mean \(\mu_{\hat p} = p\) and standard deviation \(\sigma_{\hat p} = \sqrt {\dfrac{p(1-p)}{n}}\) .

\[ \begin{cases} n = \text {sample size } \\ \hat p = \dfrac{x}{n} \text { (sample proportion) } \\ p = \text {population proportion } \\ q = 1 - p \\ \end{cases} \]

Test Statistic for Testing a Claim About a Proportion

\[ z = \dfrac{\hat p - p}{\sqrt {\dfrac{pq}{n}} } \]

The DMV claims that \(80\%\) of all drivers pass the driving test. In a survey of \(90\) teens, only \(61\) passed. Is there evidence that teen pass rates are significantly below \(80\%?\)

Let’s say, \(p\) is the true population proportion.

\[ \begin{align} \text {One-tailed test} &:\\ H_0&: p = 0.80 \\ H_A&: p < 0.80 \end{align} \]

Verify success-failure condition: \[ \begin{align} np \ge 10 \rightarrow 90 \times 0.80 \ge 10 \\ n(1-p) \ge 10 \rightarrow 90 \times (1-0.80) \ge 10 \end{align} \]

Therefore, the conditions for a normal model are met.

Now, \[ \begin{align} \hat p &= \frac {61}{90} = 0.678 \\ \\ SE(\hat p) &= \sqrt \frac{pq}{n} = \sqrt \frac{(0.80)(0.20)}{90} = 0.042 \\ \\ z &= \frac {0.678-0.80}{0.042} = -2.90 \\ \\ p\text{-value} &= P(z < -2.90) = 0.002 < 0.05 \end{align} \]

Hence, we reject \(H_0\) . Teen pass rate is significantly below population pass rate.

Under natural conditions, \(51.7\%\) of births are male. In Punjab India’s hospital \(56.9\%\) of the \(550\) births were male. Is there evidence that the proportion of male births is significantly different for this hospital?

\[ \begin{align} \text {Two-tailed test} &:\\ H_0&: p = 0.517 \\ H_A&: p \ne 0.517 \end{align} \]

Verify success-failure condition: \[ \begin{align} np \ge 10 \rightarrow 550 \times 0.517 \ge 10 \\ n(1-p) \ge 10 \rightarrow 550 \times (1-0.517) \ge 10 \end{align} \]

\[ \begin{align} \hat p &= 0.569 \\ \\ SE(\hat p) &= \sqrt \frac{pq}{n} = \sqrt \frac{(0.517)(1-0.517)}{550} = 0.0213 \\ \\ z &= \frac {0.569-0.517}{0.0213} = 2.44 \\ \\ p\text{-value} &= 2 \times P(z > 2.44) = 2 \times 0.0073 = 0.0146 < 0.05 \end{align} \]

Hence, we reject \(H_0\) . Male birth rate is significantly higher at the hospital than the natural birth rate.

8.4 \(z \text {-test}\) | Testing Hypothesis About \(\mu\) with \(\sigma\) Known

The null hypothesis claims about a population mean \(\mu\) .

\[ \begin{cases} \mu_{\bar x} = \text {population mean } \\ \sigma = \text {population standard deviation } \\ n = \text {size of the sample drawn from the population } \\ \bar x = \text{sample mean} \\ \end{cases} \]

• The sample is simple random sample.
• The population is normally distributed or \(n > 30\) .

Test Statistic for Testing a Claim About a Mean

\[ z = \dfrac{\bar x - \mu_{\bar x}}{\dfrac{\sigma}{\sqrt n}} \]

The American Automobile Association reported that the mean price of a gallon of regular gasoline in the city of Los Angeles in July 2019 was \(\$4.07\) . A recently taken simple random sample of \(50\) gas stations had an average price of \(\$4.02\) . Assume that the standard deviation of prices is \(\$0.15\) . An economist is interested in determining whether the mean price is less than \(\$4.07\) . Perform a hypothesis test at the \(\alpha = 0.05\) level of significance.

\[ \begin{align} H_0&: \mu = 4.07 \\ H_A&: \mu < 4.07 \\ \\ \bar x &= 4.02 \\ \sigma &= 0.15 \\ SE(\bar x) &= 0.15/\sqrt {50} = 0.0212 \\ \\ z &= (4.02 - 4.07)/0.0212 = -2.36 \\ p-value &= 0.09\% < 5\% \end{align} \]

Therefore, we reject \(H_0\) at the \(\alpha = 0.05\) level. We conclude that the mean price of a gallon of regular gasoline in Los Angeles is less than \(\$4.07\) .

8.5 \(t \text {-test}\) | Testing Hypothesis About \(\mu\) with \(\sigma\) Not Known

\[ \begin{cases} \mu_{\bar x} = \text {population mean } \\ s = \text {sample standard deviation } \\ n = \text {size of the sample drawn from the population } \\ \bar x = \text{sample mean} \\ \end{cases} \]

\[ t_{n-1} = \dfrac{\bar x - \mu_{\bar x}}{\dfrac{s}{\sqrt n}} \]

Average weight of a mice population of a particular breed and age is \(30 \text{ gm}\) . Weights recorded from a random sample of \(5\) mice from that population are \({31.8, 30.9, 34.2, 32.1, 28.8}.\) Test whether the sample mean is significantly greater than the population mean.

\[ \begin{align} H_0&: \mu = 30 \\ H_A&: \mu > 30 \\ \\ \bar x &= 31.56 \\ s &= 1.9604 \\ SE(\bar x) &= 1.9604/\sqrt 5 = 0.8767 \\ \\ t &= (31.56 - 30)/0.8767 = 1.779 \\ df &= (5 -1) = 4 \\ p-value &= 7.5\% > 5\% \end{align} \]

Conclusion: \(H_0\) cannot be rejected. The sample mean is not significantly greater than the population mean.

EPA recommended mirex screening is 0.08 ppm. A study of a sample of 150 salmon found an average mirex concentration of 0.0913 ppm with a std. deviation of 0.0495 ppm. Are farmed salmon contaminated beyond the permitted EPA level? Also, find a \(95\%\) confidence interval for the mirex concentration in salmon.

\[ \begin{align} H_0&: \mu = 0.08 \\ H_A&: \mu > 0.08 \\ \\ \bar x &= 0.0913 \\ s &= 0.0495 \\ SE(\bar x) &= 0.0495/\sqrt {150} = 0.0040 \\ \\ t_{149} &= \dfrac{\bar x - \mu}{SE(\bar x)} = \dfrac{(0.0913 - 0.08)}{0.0040} = 2.795 \\ df &= (150 -1) = 149 \\ p-value &= P(t_{149}>2.795)= 0.29\% < 5\% \end{align} \]

Conclusion: Reject \(H_0\) . The sample mean mirex level significantly higher that the EPA screening level.

8.6 \(\chi^2 \text{-test}\) | Testing Hypothesis About a Variance

Caution: The method of this section applies only for samples drawn from a normal distribution. If the distribution differs even slightly from normal, this method should not be used.

Listed below are the heights (cm) for the simple random sample of female supermodels. Use a \(0.01\) significance level to test the claim that supermodels have heights with a standard deviation that is less than \(\sigma=7.5 \text { cm}\) for the population of women. Does it appear that heights of supermodels vary less than heights of women from the population?

\[ \text{178, 177, 176, 174, 175, 178, 175, 178} \\ \text{178, 177, 180, 176, 180, 178, 180, 176} \\ s^2 = 3.4 \]

From \(\chi^2\) table,

\[ \text {The critical value of } \chi^2 = 5.229 \text { at } \alpha = 0.01. \\ \] Hence, we reject \(H_0\) .

Confidence Interval Calculation:

\[ \sqrt{ \dfrac{(n-1)s^2}{\chi_R^2} } < \sigma < \sqrt{ \dfrac{(n-1)s^2}{\chi_L^2} } \\ \sqrt{ \dfrac{(16-1)3.4}{30.578} } < \sigma < \sqrt{ \dfrac{(16-1)3.4}{5.229} } \\ 1.3 \text{ cm } < \sigma < 3.1 \text { cm } \]

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

Fundamental statistics for the behavioral sciences

David c howell, sampling distributions and hypothesis testing - all with video answers.

Chapter Questions

Suppose I told you that last night's NHL. hockey game resulted in a score of 26 to 13 . You would probably decide that 1 had misread the paper, because hockey games almost never have scores that high, and I was discussing something other than a hockey soore. In effect pou have just tested and rejected a null hypothesis. a) What was the null hypochesis? b) Outline the hypochesis testing procedure that you have just applied.

For the past year I have spent about $$\$ 4$$ a day for lunch, give or take a quarter or sok. a) Draw a rough sketch of this distribution of daily expenditures. b) If, without looking at the bill, I paid for my lunch with a $$\$ 5$$ bill and received $$\$ .75$$ in change, should I worry that I was crvercharged? c) Explain the logic involved in your answer to (b).

What would he a Type I error in Exercise 8.2?

What would he a Type II error in Exercise 8.2?

Using the example in Exercise 8.2, describe what we mean by the rejection region and the crirical value.

Why misht I want to adopt a one-tailed test in Exercise 8.2, and which tail should I choose? What would happen if 1 choose the wrong tail?

Using the $R$ code given for Fisure 8.1, reproduce Figure 8.2 as clooely as possible.

What would happen in Exercise 8.7 if we doubled the sise of the standard deviation? a) What happers if you change each sample's size from 10 to 100 ?

It is known that if people are asked to make an estimate of something, for example, "How tall is the University chapel?" the average guess of a group of people is more accurate than an individual's guess. Vul and Pashler (2008) wondered if the same held for multiple guesses by the same person. They asked people to make guesses about known facts. For example, "What percentage of the world's airports is in the United States?" Three weeks later the researchers asked the same people the same questions and averaged each person's responses aver the two sessions. They asked wherher this average was more accurate than the first guess by itself. We will come hack to this example later in the book. a) What are the null and altemative hypotheses? b) What would be a Type I and Type II error in this case? c) Would you be inclined to use a one-tailed or a two-tailed test in this case?

Define "simpling erroe."

What is the difference between a "distribution" and a "sampling distribution"?

How would decreasing $\alpha$ affect the probabilities given in Table 8.1!

Magen, Dweck, and Gross (2008) asked participants to chocse, for example, between $$\$ 5$$ today or $$\$ 7$$ next week. In one condition, the choices were phrased exactly that way. In a second condition, they were phrased as $$\$ 5$$ today and $$\$ 0$$ next week or $$\$ 0$$ today and $$\$ 7$$ next week," which is obviously the same thing. Each person's score was the number of choices in which the smaller but sooner choice was made. The mean for the first group was 9.24 and the mean for the second group was 6.10 . a) What are the null and alternative hypotheses? b) What statistics would you compare to answer the question? (You do not yer know how to make that comparison.) c) If the difference is significant with a two-tailed test, what would you conclude?

For the distrihution in Figure $8.6 \mathrm{I}$ suid that the probability of a Type II ermer $(\beta)$ is .64. Show how this prohahility was obtained.

Rerun the calculations in Exercise 8.14 for $\alpha=.01$.

In the example in Section 8.10, what would we have done differently if we had chosen to run a two-tailed test?

Describe the steps you would go through to flesh out the example given in this chapter about the course evaluations. In other wonds, how misht you go about determining if there truly is a relationship between grades and course evaluations?

Describe the steps you would go through to test the hypothesis that people are more likely to keep watching a movie if they have already invested money to obtain that movie.

In the exercises in Chapter 2, we discussed a study of alkowances in fourth-grade children. We considered that study again in Chapter 4 , where you generated data that might have been found in such a study. a) Consider how you would go about testing the research hypothesis that boys receive more allowance than girls. What would be the null hypothesis? b) Would you use a one-tailed or a two-tailed test? c) What results might lead you to reject the null hypothesis, and what misht lead you to retain it? d) What might you do to make this study more convincing?

Simon and Bruce (1991), in demonstrating an approach to statistics called "resampling statistics," tested the null hypothesis that the price of liquor (in 1961) for the 16 "monopoly" states, where the state owned the liquor stores, was different from the mean price in the 26 "private" states, where liquor stores were privately coned. (The means were $$\$ 4.35$$ and $$\$ 4.84$$, respectively, giving you some hint at the effects of inflation.) For technical reasons, several states don't conform to this scheme and could not be analyzed. a) What is the null hypothesis that we are actually testing? b) What label would you apply to $$\$ 4.35$$ and $$\$ 4.84$$ ? c) If these are the only states that qualify for our consideration, why are we testing a null hypothesis in the first place? d) Identify a situation in which it does make sense to test a null hypothesis here.

Several times in this chapter I have drawn a parallel between hypothesis testing and our judi. cial system. How would you describe the workings of cur judicial system in terms of Type I and Type II errors and in terms of power?