Questions: 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 86.9 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the accompanying table. The normal probability plot and boxplot for the sample are provided. Complete parts (a) and (b).
The sample was obtained using simple random sampling or from a random experiment. The boxplot does not show any outliers. The correlation coefficient from the normal probability plot is greater than the critical value, so it reasonable to conclude that the data could have come from a population that is normally distributed. It can reasonably be assumed that the sample size is less than 5% of the population size. Therefore, the requirements to perform the test using the t-distribution are satisfied.
(b) Is the new system effective? Use the α=0.01 level of significance.
State the appropriate null and alternative hypotheses.
H0: μ=86.9
H1: μ<86.9
Identify the test statistic. The test statistic is . (Round to two decimal places as needed.)
Transcript text: 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 86.9 seconds. A manager devises a new drive-through system that she believes will decrease wait time. As a test, she initiates the new system at her restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the accompanying table. The normal probability plot and boxplot for the sample are provided. Complete parts (a) and (b).
The sample was obtained using simple random sampling or from a random experiment. The boxplot does not show any outliers. The correlation coefficient from the normal probability plot is greater than the critical value, so it reasonable to conclude that the data could have come from a population that is normally distributed. It can reasonably be assumed that the sample size is less than 5% of the population size. Therefore, the requirements to perform the test using the t-distribution are satisfied.
(b) Is the new system effective? Use the α=0.01 level of significance.
State the appropriate null and alternative hypotheses.
\[
\begin{array}{l}
H_{0}: \mu=86.9 \\
H_{1}: \mu<86.9
\end{array}
\]
Identify the test statistic.
The test statistic is .
(Round to two decimal places as needed.)
Solution
Solution Steps
Step 1: Calculate the Sample Mean
The sample mean \( \bar{x} \) is calculated using the formula:
\[
\bar{x} = \frac{\sum_{i=1}^N x_i}{N} = \frac{835}{10} = 83.5
\]
Step 2: Calculate the Sample Standard Deviation
The sample standard deviation \( s \) is calculated as follows:
\[
s = \sqrt{\frac{\sum_{i=1}^N (x_i - \bar{x})^2}{N-1}} \approx 5.8
\]
Step 3: Calculate the Standard Error
The standard error \( SE \) is computed using the formula:
\[
SE = \frac{s}{\sqrt{n}} = \frac{5.8}{\sqrt{10}} \approx 1.83
\]
Step 4: Calculate the Test Statistic
The test statistic \( t \) is calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{SE} = \frac{83.5 - 86.9}{1.83} \approx -1.85
\]
Step 5: Determine the P-value
For a left-tailed test, the P-value is found to be:
\[
P = T(z) = 0.05
\]
Step 6: Conclusion
Since the P-value \( 0.05 \) is not less than the significance level \( \alpha = 0.01 \), we fail to reject the null hypothesis. Thus, there is not enough evidence to conclude that the new system is effective in reducing the wait time.
Final Answer
The correct answer is A. The test statistic is \( \boxed{-1.85} \).