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.
-
Alternative Hypothesis (H₁): The mean customer rating is different from 75 (this implies a two-tailed test).
(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: Where:
- (sample mean)
- (hypothesized population mean)
- (population standard deviation)
- (sample size)
Now, plug in the values:
The test statistic is -0.976.
(c) Conclusion
The calculated Z-test statistic is -0.976, which is within the range of , 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:
- What is the significance of using a two-tailed test in this problem?
- How would the conclusion change if we used a different level of significance?
- What would happen to the test statistic if the sample size increased?
- How does the population standard deviation affect the Z-test?
- 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
Related Recommendation
Hypothesis Testing for Customer Satisfaction Ratings - Z-Test at 0.10 Significance
Hypothesis Testing for Restaurant Wait Times Using Z-Test
Hypothesis Testing for Customer Satisfaction: Probability and Statistical Analysis
Hypothesis Testing with Customer Satisfaction Surveys: Binomial Probability Approach
Hypothesis Testing: Customer Satisfaction Rating Example