Recall that \"very satisfied\" customers give the XYZ-Box video game system a ra
ID: 3274276 • Letter: R
Question
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 67 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42.
(a) Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null hypothesis H0 and the alternative hypothesis Ha needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42.
H0: µ (Click to select)=>< 42 versus Ha: µ (Click to select)=>< 42.
z =
Reject H0 with = (Click to select).01, .001.10, .05, .01.10.001 , but not with =(Click to select).10.01, .001.001.10, .05, .01
(c) Using the information in part (b), calculate the p-value and use it to test H0 versus Ha at each of = .10, .05, .01, and .001. (Round your answers to 4 decimal places.)
(d) How much evidence is there that the mean composite satisfaction rating exceeds 42?
There is (Click to select)strongweakextremely strongvery strongno evidence.
Explanation / Answer
Solution:-
a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: < 42
Alternative hypothesis: > 42
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.3274
b)
z0.10 = 1.28
z0.05 = 1.645
z0.01 = 2.33
z0.001 = 3.09
z = (x - ) / SE
z = 2.75
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Thus, the P-value = 0.003
Interpret results.
The pvalue(0.003) is less than significance level, 0.10, 0.01, 0.05, hence we have to reject the null hypothesis.
There is very strong evidence that the mean composite satisfaction rating exceeds 42.
The pvalue(0.003) is greater than significance level, 0.001, hence we have to accept the null hypothesis.
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