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 use 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.
A) Letting m represent the mean composite satisfaction rating for the XYZ- Box, set up the null hypothesis H0 and the alternative hypothesis H a needed if we wish to attempt to provide evidence supporting the claim that m exceeds 42.
B) The random sample of 65 satisfaction ratings yields a sample mean of x = 42.954 . Assuming that o equals 2.64, use critical values to test H0 versus H a at each of the following .10, .05, .01, and .001.
C) Using the information in part b, calculate the p- value and use it to test H0 versus H a at each of the following .10, .05, .01, and .001.
D) How much evidence is there that the mean composite satisfaction rating exceeds 42?
A) The null hypothesis (H0) in this case would be that the mean composite satisfaction rating for the XYZ-Box is not greater than 42. The alternative hypothesis (Ha) would be that the mean composite satisfaction rating for the XYZ-Box exceeds 42.
B) To test the null hypothesis, we need to calculate the test statistic using the sample mean, population standard deviation (o), and critical values. The critical values depend on the chosen significance level (alpha).
Let's assume alpha (α) = 0.10:
The critical value for a one-tailed test with alpha = 0.10 and a sample size of 65 can be found using a statistical table or calculator. Let's assume it is 1.28.
The test statistic (z) can be calculated using the formula: z = (x - m) / (o / sqrt(n))
Where:
x = sample mean (42.954)
m = hypothesized population mean (42, since H0 states it is not greater than 42)
o = population standard deviation (2.64)
n = sample size (65)
Substituting the given values into the formula:
z = (42.954 - 42) / (2.64 / sqrt(65))
z ≈ 0.529
Since the calculated test statistic (0.529) is less than the critical value (1.28), we fail to reject the null hypothesis.
Repeat the calculation for alpha levels of 0.05, 0.01, and 0.001 to determine if the null hypothesis can be rejected at those significance levels.
C) To calculate the p-value, we need to find the area under the standard normal distribution curve corresponding to the test statistic (z-score). The p-value represents the probability of obtaining a sample mean as extreme or more extreme than the one observed, assuming the null hypothesis is true.
Using the Z-table or a calculator, we find that the area to the left of the z-score 0.529 is approximately 0.7009. Subtracting this value from 1 gives us the p-value.
For alpha = 0.10:
p-value = 1 - 0.7009 ≈ 0.2991
Repeat the calculation for alpha levels of 0.05, 0.01, and 0.001 to determine the corresponding p-values.
D) The evidence provided by the data depends on the level of significance (alpha) chosen. If the evidence suggests rejecting the null hypothesis, it indicates that there is some evidence that the mean composite satisfaction rating exceeds 42. However, if the evidence is not significant enough (p-value greater than alpha), we fail to reject the null hypothesis, suggesting insufficient evidence to support that the mean composite satisfaction rating exceeds 42.