Questions: The data table contains waiting times of customers at a bank, where customers enter a single waiting line that feeds three teller windows. Test the claim that the standard deviation of waiting times is less than 1.1 minutes, which is the standard deviation of waiting times at the same bank when separate waiting lines are used at each teller window. Use a significance level of 0.01. Assume that the sample is a simple random sample selected from a normally distributed population. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. A. H0: σ<1.1 minutes B. H0: σ ≥ 1 HA: σ< H0: σ= HA: σ< c. H0: σ=1.1 minutes HA: σ ≠ 1.1 minutes b. Compute the test statistic. x^2= (Round to two decimal places as needed.) Customer Waiting Times Customer Waiting Times (in minutes) 6.8 6.9 5.6 6.7 6.6 6.2 6.1 6.2 7.6 7.6 6.8 7.1 7.6 7.4 5.3 7.9 8.7 6.1 7.1 6.7 7.7 7.5 6.4 6.1 7.3 6.6 6.8 8.2 7.9 6.7 6.5 6.1 72 8.4 6.1 7.3 6.4 6.2 8.3 6.1 6.7 6.5 8.3 8.3 8.6 7.6 7.5 7.2 56 6.7 7.6 7.7 6.5 7.3 7.5 7.8 6.1 7.8 7.3 7.5

The data table contains waiting times of customers at a bank, where customers enter a single waiting line that feeds three teller windows. Test the claim that the standard deviation of waiting times is less than 1.1 minutes, which is the standard deviation of waiting times at the same bank when separate waiting lines are used at each teller window. Use a significance level of 0.01. Assume that the sample is a simple random sample selected from a normally distributed population. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. A. H0: σ<1.1 minutes B. H0: σ ≥ 1 HA: σ< H0: σ= HA: σ< c. H0: σ=1.1 minutes HA: σ ≠ 1.1 minutes b. Compute the test statistic. x^2= (Round to two decimal places as needed.) Customer Waiting Times Customer Waiting Times (in minutes) 6.8 6.9 5.6 6.7 6.6 6.2 6.1 6.2 7.6 7.6 6.8 7.1 7.6 7.4 5.3 7.9 8.7 6.1 7.1 6.7 7.7 7.5 6.4 6.1 7.3 6.6 6.8 8.2 7.9 6.7 6.5 6.1 72 8.4 6.1 7.3 6.4 6.2 8.3 6.1 6.7 6.5 8.3 8.3 8.6 7.6 7.5 7.2 56 6.7 7.6 7.7 6.5 7.3 7.5 7.8 6.1 7.8 7.3 7.5
Transcript text: The data table contains waiting times of customers at a bank, where customers enter a single waiting line that feeds three teller windows. Test the claim that the standard deviation of waiting times is less than 1.1 minutes, which is the standard deviation of waiting times at the same bank when separate waiting lines are used at each teller window. Use a significance level of 0.01. Assume that the sample is a simple random sample selected from a normally distributed population. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. A. $\mathrm{H}_{0}: \sigma<1.1$ minutes B. $H_{0}: \sigma \geq 1$ $H_{A}: \sigma<$ $H_{0}: \sigma=$ $H_{A}: \sigma<$ c. $\mathrm{H}_{0}: \sigma=1.1$ minutes $H_{A}: \sigma \neq 1.1$ minutes b. Compute the test statistic. \[ x^{2}=\square \] (Round to two decimal places as needed.) Customer Waiting Times \begin{tabular}{|c|c|c|c|c|} \hline \multirow[t]{16}{*}{} & \multicolumn{4}{|c|}{Customer Waiting Times (in minutes)} \\ \hline & 6.8 & 6.9 & 5.6 & 6.7 \\ \hline & 6.6 & 6.2 & 6.1 & 6.2 \\ \hline & 7.6 & 7.6 & 6.8 & 7.1 \\ \hline & 7.6 & 7.4 & 5.3 & 7.9 \\ \hline & 8.7 & 6.1 & 7.1 & 6.7 \\ \hline & 7.7 & 7.5 & 6.4 & 6.1 \\ \hline & 7.3 & 6.6 & 6.8 & 8.2 \\ \hline & 7.9 & 6.7 & 6.5 & 6.1 \\ \hline & 72 & 8.4 & 6.1 & 7.3 \\ \hline & 6.4 & 6.2 & 8.3 & 6.1 \\ \hline & 6.7 & 6.5 & 8.3 & 8.3 \\ \hline & 8.6 & 7.6 & 7.5 & 7.2 \\ \hline & 56 & 6.7 & 7.6 & 7.7 \\ \hline & 6.5 & 7.3 & 7.5 & 7.8 \\ \hline & 6.1 & 7.8 & 7.3 & 7.5 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Calculate the Sample Mean

The sample mean \( \mu \) is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{538.1}{60} \approx 8.9683 \]

Step 2: Calculate the Sample Variance and Standard Deviation

The sample variance \( \sigma^2 \) is computed using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 108.98 \]

The sample standard deviation \( \sigma \) is then:

\[ \sigma = \sqrt{108.98} \approx 10.44 \]

Step 3: Calculate the Test Statistic

The test statistic \( \chi^2 \) is calculated using the formula:

\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \]

Substituting the values:

\[ \chi^2 = \frac{(60 - 1) \cdot 108.98}{1.21} \approx 5313.9 \]

Step 4: Determine the P-value

To find the P-value corresponding to the calculated \( \chi^2 \) test statistic with 59 degrees of freedom, we evaluate:

\[ P = 1.0 \]

Final Answer

The test statistic is approximately \( \chi^2 \approx 5313.9 \) and the P-value is \( P = 1.0 \).

Thus, the final answer is:

\[ \boxed{\chi^2 \approx 5313.9} \]

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