91Ó°ÊÓ

A quality-control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determine if a new loading/unloading procedure is effective in reducing wait time in line, he measures the amount of time (in minutes) people are waiting in line for seven days. After implementing the new procedure, he again measures the amount of time (in minutes) people are waiting in line for seven days and obtains the data on the next page. To make a reasonable comparison, he chooses days when weather conditions are alike. Treat each day as a block and the wait times before and after the procedure as the treatment. $$ \begin{array}{lccccccc} \text { Day } & \text { Mon } & \text { Tues } & \text { Wed } & \text { Thurs } & \text { Fri } & \text { Sat } & \text { Sun } \\ \hline \begin{array}{l} \text { Wait time before } \\ \text { new procedure } \end{array} & 11.6 & 25.9 & 20.0 & 38.2 & 57.3 & 32.1 & 81.8 \\ \hline \begin{array}{l} \text { Wait time after } \\ \text { new procedure } \end{array} & 10.7 & 28.3 & 19.2 & 35.9 & 59.2 & 31.8 & 75.3 \\ \hline \end{array} $$ (a) Using the methods introduced in this section, determine whether there is sufficient evidence to conclude that the two loading procedures are resulting in different measurements of the wait time at the \(\alpha=0.05\) level of significance. (b) Using the methods introduced in Section \(11.2,\) determine whether there is sufficient evidence to conclude that the two loading procedures are resulting in different measurements of the wait time at the \(\alpha=0.05\) level of significance. (c) Compare the \(P\) -values of both approaches. Can you conclude that the method presented in this section is a generalization of the matched-pairs \(t\) -test?

Step by step solution

01

State the Hypotheses

For both parts (a) and (b), we need to state the null and alternative hypotheses.For part (a):- Null Hypothesis (H_{0}): The amount of wait time before and after the new procedure is the same.- Alternative Hypothesis (H_{1}): The amount of wait time before and after the new procedure is different.For part (b):- Null Hypothesis (H_{0}): The mean difference in wait times before and after the new procedure is zero.- Alternative Hypothesis (H_{1}): The mean difference in wait times before and after the new procedure is not zero.

02

Calculate the Differences (Part a)

Calculate the differences between the wait times before and after the new procedure for each day.\[\begin{array}{lccccccc} \text { Day } & \text { Mon } & \text { Tues } & \text { Wed } & \text { Thurs } & \text { Fri } & \text { Sat } & \text { Sun } \ \hline \text { Difference } & 0.9 & -2.4 & 0.8 & 2.3 & -1.9 & 0.3 & 6.5 \ \end{array} \]

03

Conduct the Matched-Pairs t-Test (Part a)

Compute the mean (\overline{d}), standard deviation (\sigma_{d}) of the differences, and the test statistic. Mean difference (\overline{d}):\[\overline{d} = \frac{0.9 + -2.4 + 0.8 + 2.3 + -1.9 + 0.3 + 6.5}{7} = 0.7857\]Standard deviation of the differences (\sigma_{d}):Use the formula \(\sigma_{d} = \sqrt{\frac{\sum (x - \overline{d})^{2}}{n - 1}}\)Test statistic (t):\[t = \frac{\overline{d}}{\sigma_{d}/\sqrt{n}}\]

04

Compute the Test Statistic (Part b)

Using ANOVA for blocked designs, we need to calculate the test statistic. First compute the sum of squares for treatments (SST), the sum of squares for blocks (SSB), and the sum of squares for error (SSE). Use the test statistic:\[F = \frac{MS_{T}}{MS_{E}}\]Where \(MS_{T}\) is the mean square for treatments, and \(MS_{E}\) is the mean square for error.

05

Determine the Critical Values and Make Decisions

For both parts, compare the calculated test statistics to their respective critical values from t-distribution and F-distribution tables at \(\alpha = 0.05\). If the test statistic falls in the rejection region, reject the null hypothesis.

06

Compare P-values

Calculate the P-values for both tests. If the P-values are less than \(\alpha = 0.05\), the null hypothesis is rejected.Compare the P-values of both approaches to determine if ANOVA is a generalization of the matched-pairs t-test.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

P-Values Comparison

P-values tell us the probability of obtaining the observed data, or something even more extreme, if the null hypothesis is true. Comparing p-values from different statistical tests helps understand their strengths and limitations.

For both the matched-pairs t-test and the ANOVA for blocked designs, once we calculate the test statistics, we determine their p-values:

• For the t-test, we use the t-distribution to find the p-value corresponding to the calculated t-statistic.
• For ANOVA, we use the F-distribution to find the p-value associated with the computed F-statistic.

If the p-values are less than 0.05 (our significance level), we reject the null hypothesis, indicating that the new procedure has significantly different wait times.

By comparing p-values from both methods, you can see whether ANOVA, using more comprehensive calculations accounting for blocks, agrees with the simpler matched-pairs t-test. Typically, if ANOVA shows significant results, the t-test result is a specific case, implying that ANOVA generalizes the matched-pairs method.