Advanced High School Statistics: First Edition

The first difference shown in Table 7.2.2 is computed as \(27.67-27.95=-0.28\text{.}\) Based on the table and on the histogram of differences in Figure 7.2.3, which store tends to have the higher prices in the sample?²Because the most of the differences are positive, UCLA prices tend to be higher than Amazon. Note that it is important to identify the order in which the differences are taken.

Set up and implement a hypothesis test to determine whether, on average, there is a difference between Amazon's price for a book and the UCLA bookstore's price.

Solution

There are two scenarios: there is no difference or there is some difference in average prices. The no difference scenario is always the null hypothesis:

• \(\mu_{diff}=0\text{.}\) There is no difference in the average textbook price.

• \(\mu_{diff} \neq 0\text{.}\) There is a difference in average prices.

The standard deviation of all of the differences in unknown, so we will use the standard deviation of the sample differences. The observations are based on a simple random sample from less than 10% of all books sold at the bookstore, so independence is reasonable; the distribution of differences, shown in Figure 7.2.3, is strongly skewed, but the sample size \(n=73\) is well over 30. Because all three conditions are reasonably satisfied, we can conclude the \(t\)-test is reasonable.

We compute the standard error associated with \(\bar{x}_{diff}\) using the standard deviation of the differences (\(s_{_{diff}}=14.26\)) and the number of differences (\(n_{_{diff}}=73\)):

\begin{equation*} SE_{\bar{x}_{diff}} = \frac{s_{diff}}{\sqrt{n_{diff}}} = \frac{14.26}{\sqrt{73}} = 1.67 \end{equation*}

To visualize the p-value, the sampling distribution of \(\bar{x}_{diff}\) is drawn as though \(H_0\) is true, which is shown in Figure 7.2.7. The p-value is represented by the two (very) small tails.

To find the tail areas, we compute the test statistic, which is the \(T\) score of \(\bar{x}_{diff}\) under the null condition that the actual mean difference is 0:

\begin{equation*} t = \frac{\bar{x}_{diff} - 0}{SE_{x_{diff}}} = \frac{12.76 - 0}{1.67} = 7.65 df=72 \end{equation*}

This \(T\) score is so large it isn't even in the table, which ensures the single tail area will be 0.0002 or smaller. A calculator gives a tail area as \(4.5\times 10^{-11}\text{.}\) Since the p-value corresponds to both tails in this case and the \(t\)-distribution is symmetric, the p-value can be estimated as twice the one-tail area:

\begin{equation*} \text{ p-value } = 2\times (\text{ one tail area } ) \approx 2\times 4.5\times 10^{-11} = 9\times 10^{-11}\approx 0 \end{equation*}

Because the p-value is less than 0.05, we reject the null hypothesis. We have found convincing evidence that Amazon is, on average, cheaper than the UCLA bookstore for UCLA course textbooks.

An SAT preparation company claims that its students' scores improve by over 100 points on average after their course. A consumer group would like to evaluate this claim, and they collect data on a random sample of 30 students who took the class. Each of these students took the SAT before and after taking the company's course, and so we have a difference in scores for each student. We will examine these differences \(x_1=57\text{,}\) \(x_2=133\text{,}\) ..., \(x_{30}=140\) as a sample to evaluate the company's claim. The distribution of the differences, shown in Figure 7.2.8, has mean 135.9 and standard deviation 82.2. Do these data provide convincing evidence to back up the company's claim? ³These are paired data, so we analyze the score differences with a matched pairs \(t\)-test. Conditions: This is a random sample from less than 10% of the company's students (assuming they have more than 300 former students), so the independence condition is reasonable. \(n=30\ge 30\text{.}\) This is a one-sided test. \(H_0\text{:}\) student scores do not improve by more than 100 after taking the company's course. \(\mu_{_{diff}} = 100\) \(H_A\text{:}\) students scores improve by more than 100 points on average after taking the company's course. \(\mu_{_{diff}} \gt 100\text{.}\) Let

Because we rejected the null hypothesis, does this mean that taking the company's class improves student scores by more than 100 points on average?⁴This is an observational study, so we cannot make this causal conclusion. For instance, maybe SAT test takers tend to improve their score over time even if they don't take a special SAT class, or perhaps only the most motivated students take such SAT courses.

In the SAT preparation company example, we saw that \(\bar{x}_{diff}\) was 135.9 and \(s_{diff}\) was 82.2. That is, the average change in students' scores after the class was a 135.9 point increase and the SD of the change or difference in their scores was 82.2 points. Construct a 95% confidence interval to estimate the true average change in score after taking the class. Is there evidence for the company's claim that students score an average of 100 points higher after the class?⁵Because this is a before and after scenario, we use a matched pairs \(t\)-interval. The conditions were verified in the previous section. The confidence interval is : \(135.9\ \pm 2.045(15.0) \to (105.2, 166.6)\text{.}\) We can be 95% confident that the true average increase in scores after the prep class is between 105.2 and 166.6. Because the entire interval is above 100, there is evidence that on average students score more than 100 points higher after the course. Recall that this does not prove that the increase is due to the course.

The 95% confidence interval in the previous exercise was calculated as (105.2, 166.6). True or false: about 95% of the students that take the class saw an increase of at least 105.2 points.⁶False. This confidence interval estimates the average increase - not the increase of individuals. As can be seen in Figure 7.2.8, much greater than 5% saw an increase of less than 105.2 points. Some individuals even saw a decrease in their score as indicated by the negative differences.

Use the first 7 values of the Table 7.2.2 data set produced above and calculate the \(T\) score and p-value to test whether, on average, Amazon's textbook price is cheaper that UCLA's price.⁷Create a list of the differences, and use the data or list option to perform the test. Let \(\mu_0\) be 0, and select the appropriate list. Freq should be 1, and the test sidedness should be \(\gt\text{.}\) \(T=3.076\) and p-value\(=0.0109\text{.}\)

Use the data from Table 7.2.14 to calculate a 95% confidence interval for the average difference in textbook price between Amazon and UCLA.⁸Choose a C-Level of 0.95, and the final result should be (0.80354, 7.0507).