Question 7, 10.3.21-T
HW Score: $41.41 \%, 3.73$ of 9
Part 3 of 5 points
(2) Points: 0.73 of 1
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The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 87.1 seconds. A manager devises a new drive-through system that he believes will decrease wait time. As a test, he initiates the new system at his restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below.
\begin{tabular}{|l|l|}
\hline 102.3 & 82.2 \\
69.7 & 95.7 \\
56.3 & 86.5 \\
74.3 & 72.9 \\
64.7 & 85.4 \\
\hline
\end{tabular}
Click the icon to view the table of correlation coefficient critical values.
(a) Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers. The normal probability plot is shown below and the sample correlation coefficient is known to be $r=0.994$. Are the conditions for testing the hypothesis satisfied?
Yes, the conditions are satisfied. The normal probability plot is linear enough, since the correlation coefficient is greater than the critical value. In addition, a boxplot does not show any outliers.
(b) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of $\alpha=0.1$.
First determine the appropriate hypotheses.
\[
\begin{array}{l}
H_{0}: \mu=87.1 \\
H_{1}: \mu< 87.1
\end{array}
\]
Find the test statistic.
\[
\mathrm{t}_{0}=\square
\]
(Round to two decimal places as needed.)
Answer
Since the P-value (0.0384) is less than the level of significance (\(\alpha=0.1\)), we reject the null hypothesis. Therefore, the new system is effective in reducing the wait time. \(\boxed{\text{Reject } H_0}\)
Step 1 ：Calculate the sample mean (\(\bar{x}\)) as the sum of all the sample values divided by the number of samples: \(\bar{x} = \frac{102.3 + 82.2 + 69.7 + 95.7 + 56.3 + 86.5 + 74.3 + 72.9 + 64.7 + 85.4}{10} = 79.0\)
Step 2 ：Calculate the sample standard deviation (s) as the square root of the sum of the squared differences between each sample value and the sample mean, divided by the number of samples minus 1: \(s = \sqrt{\frac{(102.3-79.0)^2 + (82.2-79.0)^2 + (69.7-79.0)^2 + (95.7-79.0)^2 + (56.3-79.0)^2 + (86.5-79.0)^2 + (74.3-79.0)^2 + (72.9-79.0)^2 + (64.7-79.0)^2 + (85.4-79.0)^2}{10-1}} = 12.8\)
Step 3 ：Calculate the test statistic (\(t_0\)) as the difference between the sample mean and the population mean, divided by the standard error of the mean. The standard error of the mean is the sample standard deviation divided by the square root of the number of samples: \(t_0 = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{79.0 - 87.1}{12.8 / \sqrt{10}} = -2.00\)
Step 4 ：Find the P-value associated with this test statistic. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. Using a t-distribution table or a statistical software, we find that the P-value associated with a t-statistic of -2.00 with 9 degrees of freedom is 0.0384.
Step 5 ：Since the P-value (0.0384) is less than the level of significance (\(\alpha=0.1\)), we reject the null hypothesis. Therefore, the new system is effective in reducing the wait time. \(\boxed{\text{Reject } H_0}\)
