Recall that “very satisfied” customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to us the random sample of 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42.

Need mean and standard deviation of sample. What level of significance (P = .05, P = .01) are you using?

To provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42, the manufacturer needs to conduct a hypothesis test. Here's how they can do it:

1. State the null and alternative hypotheses:
- Null hypothesis (H0): The mean composite satisfaction rating for the XYZ-Box is not greater than 42.
- Alternative hypothesis (Ha): The mean composite satisfaction rating for the XYZ-Box exceeds 42.

2. Determine the significance level (α) for the test. This is the maximum probability of committing a Type I error, which is rejecting the null hypothesis when it is true. Commonly used values are 0.05 or 0.01.

3. Select the appropriate statistical test. Since we are comparing the mean of a single sample to a specific value, a one-sample t-test is appropriate in this case.

4. Collect the data. The manufacturer needs to obtain the random sample of 65 satisfaction ratings from customers.

5. Calculate the sample mean and standard deviation of the satisfaction ratings.

6. Conduct the hypothesis test:
- Calculate the test statistic, which in this case is the t-statistic. The formula for the t-statistic is:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
- Determine the degrees of freedom for the t-distribution, which is equal to sample size minus 1.
- Use the calculated t-statistic and degrees of freedom to find the p-value associated with the test statistic. The p-value represents the probability of observing a sample mean at least as extreme as the one obtained, assuming the null hypothesis is true.
- Compare the p-value with the significance level (α). If the p-value is less than α, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

7. Interpret the results. If the null hypothesis is rejected, there is evidence to support the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42. If the null hypothesis is not rejected, there is not enough evidence to support the claim.

Note that conducting the hypothesis test requires statistical software or a calculator that can perform t-tests or provide the necessary calculations.