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A chain of restaurants has historically had a mean wait time of 9 minutes for its customers. Recently, the restaurant added several very popular dishes back to their menu. Due to this, the manager suspects the wait time, \(\mu\), has increased. He takes a random sample of 44 customers. The mean wait time for the sample is 10.4 minutes. Assume, the population standard deviation for the wait times is known to be 4.1 minutes. Can the manager conclude that the mean wait time is now greater than 9 minutes? Perform a hypothesis test, using the 0.01 level of significance. (a) State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\). \(H_0:\ \) \(H_1:\ \) (b) Perform a Z-test and find the p-value. Here is some information to help you with your Z-test. \(ullet\) The value of the test statistic is given by \(\displaystyle Z = rac{\overline{x}-\mu}{\sigma/\sqrt{n}}\) \(ullet\) The p-value is the area under the curve to the right of the value of the test statistic. \(ullet\) <Standard Normal Distribution Table> \(\mu=0,\sigma=1\) Step 1: Compute the test statistic. Step 2: Find the p-value. Step 3: Shade the area represented by the p-value. Step 4: Enter the p-value. (Round to 3 decimal places.) (c) Based on your answer to part (b), choose what the manager can conclude, at the 0.01 level of significance. \(\square\) Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes. \(\square\) Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes. \(\square\) Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes. \(\square\) Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes. <The image contains a bell curve with shaded areas and a Standard Normal Distribution table.>

          A chain of restaurants has historically had a mean wait time of 9 minutes for its customers. Recently, the restaurant added several very popular dishes back to their menu. Due to this, the manager suspects the wait time, \(\mu\), has increased. He takes a random sample of 44 customers. The mean wait time for the sample is 10.4 minutes. Assume, the population standard deviation for the wait times is known to be 4.1 minutes.

Can the manager conclude that the mean wait time is now greater than 9 minutes? Perform a hypothesis test, using the 0.01 level of significance.

(a) State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\).
   \(H_0:\ \)
   \(H_1:\ \)
(b) Perform a Z-test and find the p-value.
   Here is some information to help you with your Z-test.
   \(ullet\) The value of the test statistic is given by \(\displaystyle Z = rac{\overline{x}-\mu}{\sigma/\sqrt{n}}\)
   \(ullet\) The p-value is the area under the curve to the right of the value of the test statistic.

   \(ullet\) <Standard Normal Distribution Table>
      \(\mu=0,\sigma=1\)
      Step 1: Compute the test statistic.
      Step 2: Find the p-value.
      Step 3: Shade the area represented by the p-value.
      Step 4: Enter the p-value. (Round to 3 decimal places.)

(c) Based on your answer to part (b), choose what the manager can conclude, at the 0.01 level of significance.
      \(\square\) Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes.
      \(\square\) Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes.
      \(\square\) Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes.
      \(\square\) Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes.

<The image contains a bell curve with shaded areas and a Standard Normal Distribution table.>

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A chain of restaurants has historically had a mean wait time of 9 minutes for its customers. Recently, the restaurant added several very popular dishes back to their menu. Due to this, the manager suspects the wait time, μ, has increased. He takes a random sample of 44 customers. The mean wait time for the sample is 10.4 minutes. Assume, the population standard deviation for the wait times is known to be 4.1 minutes.

Can the manager conclude that the mean wait time is now greater than 9 minutes? Perform a hypothesis test, using the 0.01 level of significance.

(a) State the null hypothesis H0 and the alternative hypothesis H1.
H0:
H1:
(b) Perform a Z-test and find the p-value.
Here is some information to help you with your Z-test.
ullet The value of the test statistic is given by Z = racx-μσ/√(n)
ullet The p-value is the area under the curve to the right of the value of the test statistic.

ullet <Standard Normal Distribution Table>
μ=0,σ=1
Step 1: Compute the test statistic.
Step 2: Find the p-value.
Step 3: Shade the area represented by the p-value.
Step 4: Enter the p-value. (Round to 3 decimal places.)

(c) Based on your answer to part (b), choose what the manager can conclude, at the 0.01 level of significance.
□ Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes.
□ Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes.
□ Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean wait time is now greater than 9 minutes.
□ Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to conclude that the mean wait time is now greater than 9 minutes.

<The image contains a bell curve with shaded areas and a Standard Normal Distribution table.>

Added by Vanessa L.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Pritesh Ranjan

Step 1

\(H_0: \mu = 9\) minutes (The mean wait time has not increased). \(H_1: \mu > 9\) minutes (The mean wait time has increased). Show more…

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ransfe... Gmail 3.8 The Impact of... Precalculus Functi... \( -1 \) \( -2 \) r 3 \( =4 \) \( -5 \) 6 7 (8) A chain ef restarints has historicaly had a mean wat time of 9 mimutes for its customers. Recently, the restmrant added severat very poputur diuhes badk to their menu. Due to mis, the manajer suspects tha wat time, H, has inkreased. He takes a random sample of 44 cost \[ \begin{array}{l} \boldsymbol{H}_{4}: \square \\ H_{1}: \square \end{array} \] \( \left[\begin{array}{ccc}\mu & \pi & C^{p} \\ 0 \times 0 & 0 \times 0 & 0.0 \\ 0 \times 0 & 0+0 & 0 \times 0 \\ x & 3\end{array}\right] \) (b) Perform a Z-test and find the p-value. Here is some intorribrion to hep you with your \( Z_{\text {test }} \). * The value of the test statistic is given by \( \frac{\bar{x}-\mu}{\frac{0}{\sqrt{n}}} \). - The pvalus is the mese under the couva to the ribbte of the value of the test statistic. \( \times \quad 5 \) exidence to ennchite that the men wak time is new grester than 9 minutes. moweh svifence te coseive that the mean wet is now preater than 9 minetes. to cenchate that the meal wall tine is now preater than 9 minutes. evidence ta Escluse that the mean war tige is nor preater than 9 mandes.

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