Math Problem Statement

Skip to Main Content Español

Topic 5 Homework (Nonadaptive) Question 6 of 21 (1 point)|Question Attempt: 1 of Unlimited

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Question 6 Over the years, the mean customer satisfaction rating at a local restaurant has been 75. The restaurant was recently remodeled, and now the management claims the mean customer rating, μ, is not equal to 75. In a sample of 37 customers chosen at random, the mean customer rating is 72.5. Assume that the population standard deviation of customer ratings is 15.6.

Is there enough evidence to support the claim that the mean customer rating is different from 75? Perform a hypothesis test, using the 0.10 level of significance.

(a) State the null hypothesis H0 and the alternative hypothesis H1.

H0:

H1:

(b)Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your test. z0.05 is the value that cuts off an area of 0.05 in the right tail. The test statistic has a normal distribution and the value is given by =z−xμσn. Standard Normal Distribution Step 1: Select one-tailed or two-tailed. One-tailed Two-tailed Step 2: Enter the critical value(s). (Round to 3 decimal places.)

Step 3: Enter the test statistic. (Round to 3 decimal places.)

0.10.20.30.41−12−23−3

(c)Based on your answer to part (b), choose what can be concluded, at the 0.10 level of significance, about the claim made by the management. Conclusion:

Since the value of the test statistic lies in the rejection region, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal 75.

Since the value of the test statistic lies in the rejection region, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean customer rating is not equal 75.

Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean customer rating is not equal 75.

Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean customer rating is not equal 75.

Check

Save For Later

Submit Assignment © 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility Clears your work. Undoes your last action.

Solution

Let's walk through the solution step by step. We're conducting a hypothesis test on the mean customer rating of a restaurant. Here's how to approach this:

(a) Null Hypothesis and Alternative Hypothesis

We are testing whether the mean customer rating has changed after the restaurant was remodeled. The hypotheses are:

  • Null Hypothesis (H₀): The mean customer rating is still 75, meaning the remodeling had no effect. H0:μ=75H₀: \mu = 75

  • Alternative Hypothesis (H₁): The mean customer rating is different from 75 (this implies a two-tailed test). H1:μ75H₁: \mu \neq 75

(b) Performing the Hypothesis Test (Z-test)

Step 1: Type of test

Since we are testing whether the mean is different from 75 (either higher or lower), this is a two-tailed test.

Step 2: Calculate the critical value

At the 0.10 significance level for a two-tailed test, the significance level in each tail is 0.05. The critical values for a Z-test at the 0.05 level are approximately ±1.645. These are the critical Z-values that define the rejection regions.

Step 3: Calculate the test statistic

We use the formula for the Z-test: Z=xˉμσnZ = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} Where:

  • xˉ=72.5\bar{x} = 72.5 (sample mean)
  • μ=75\mu = 75 (hypothesized population mean)
  • σ=15.6\sigma = 15.6 (population standard deviation)
  • n=37n = 37 (sample size)

Now, plug in the values: Z=72.57515.637=2.52.5620.976Z = \frac{72.5 - 75}{\frac{15.6}{\sqrt{37}}} = \frac{-2.5}{2.562} \approx -0.976

The test statistic is -0.976.

(c) Conclusion

The calculated Z-test statistic is -0.976, which is within the range of [1.645,1.645][-1.645, 1.645], meaning it does not fall in the rejection region.

Thus, the correct conclusion is:

  • Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is not rejected. Therefore, there is not enough evidence to support the claim that the mean customer rating is different from 75.

Do you want more details or have any questions?

Here are five questions to further expand your understanding:

  1. What is the significance of using a two-tailed test in this problem?
  2. How would the conclusion change if we used a different level of significance?
  3. What would happen to the test statistic if the sample size increased?
  4. How does the population standard deviation affect the Z-test?
  5. Why is it important to check whether the population standard deviation is known?

Tip: In hypothesis testing, choosing the correct tail for the test (one-tailed or two-tailed) is essential based on the alternative hypothesis.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Z-Test
Statistical Significance

Formulas

Z = (x̄ - μ) / (σ / √n)
Critical Z-value for two-tailed test at α = 0.10

Theorems

Central Limit Theorem
Properties of Z-distribution

Suitable Grade Level

Undergraduate Statistics