Questions: Over the years, the mean customer satisfaction rating at a local restaurant has been 80. The restaurant was recently remodeled, and now the management claims the mean customer rating, μ, is not equal to 80. In a sample of 54 customers chosen at random, the mean customer rating is 75.4. Assume that the population standard deviation of customer ratings is 21.2 Is there enough evidence to support the claim that the mean customer rating is different from 80? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis H0 and the alternative hypothesis H1 (b) Perform a Z-test and find the p-value Here is some information to help you with your Z-test. - The value of the test statistic is given by (x̄-μ)/(σ/√n).

Over the years, the mean customer satisfaction rating at a local restaurant has been 80. The restaurant was recently remodeled, and now the management claims the mean customer rating, μ, is not equal to 80. In a sample of 54 customers chosen at random, the mean customer rating is 75.4. Assume that the population standard deviation of customer ratings is 21.2 Is there enough evidence to support the claim that the mean customer rating is different from 80? Perform a hypothesis test, using the 0.05 level of significance.
(a) State the null hypothesis H0 and the alternative hypothesis H1
(b) Perform a Z-test and find the p-value Here is some information to help you with your Z-test.
- The value of the test statistic is given by (x̄-μ)/(σ/√n).

Transcript text: Over the years, the mean customer satisfaction rating at a local restaurant has been 80. The restaurant was recently remodeled, and now the management claims the mean customer rating, $\mu$, is not equal to 80. In a sample of 54 customers chosen at random, the mean customer rating is 75.4. Assume that the population standard deviation of customer ratings is 21.2 Is there enough evidence to support the claim that the mean customer rating is different from 80? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ (b) Perform a $Z$-test and find the $p$-value Here is some information to help you with your $Z$-test. - The value of the test statistic is given by $\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$.

Solution

Solution Steps

Step 1: State the Hypotheses

The null hypothesis $ H_0 $ and the alternative hypothesis $ H_1 $ are defined as follows:

$ H_0: \mu = 80 $ (The mean customer satisfaction rating is equal to 80)
$ H_1: \mu \neq 80 $ (The mean customer satisfaction rating is not equal to 80)

Step 2: Calculate the Standard Error

The standard error $ SE $ is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{21.2}{\sqrt{54}} \approx 2.885 \]

Step 3: Calculate the Test Statistic

The test statistic $ Z $ is calculated using the formula: \[ Z = \frac{\bar{x} - \mu_0}{SE} = \frac{75.4 - 80}{2.885} \approx -1.5945 \]

Step 4: Calculate the P-value

For a two-tailed test, the P-value is calculated as: \[ P = 2 \times (1 - T(|z|)) \approx 0.1108 \]

Step 5: Decision Rule

Using a significance level of $ \alpha = 0.05 $:

If $ P < \alpha $, reject $ H_0 $.
If $ P \geq \alpha $, fail to reject $ H_0 $.

In this case, since $ P \approx 0.1108 $ is greater than $ 0.05 $, we fail to reject the null hypothesis.

Final Answer

There is not enough evidence to support the claim that the mean customer rating is different from 80. Thus, the conclusion is: \[ \boxed{\text{Fail to reject } H_0} \]

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >

Thank you for your feedback.

Copied!