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Statistics for Managers
Running head: WEEK 4 ASSIGNMENT 1
Week 4 Assignment
Sarah Doppelmayr
Statistics for Managers
BUS308
Edward Kaplan
January 28, 2013
WEEK 4 ASSIGNMENT 2
Week 4 Assignment
Chapter 9: 9.13 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 mew 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 mew exceeds 42. H0: mu 42 B. The random sample of 65 satisfaction ratings yields a sample mean of x bar = 42.954. Assuming that sigma equals 2.64, use critical values to test H0 versus Ha at each of .10, .05, .01 and .001. z-statistic: z = (xbar - μ)/(σ/√n) z = (42.954 - 42 )/(2.64/√65) z = 0.954 / (2.64/8.0623) z = 2.9134 alpha z-crit result 0.10 1.282 significant 0.05 1.645 significant
WEEK 4 ASSIGNMENT 3
0.01 2.326 significant 0.001 3.09 not significant 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. The p-value is 0.0025. This being stated, the result is significant at 0.10, 0.05, and 0.01, because p < alpha. At 0.001, the result is not significant, because p > alpha How much evidence is there that the mean composite satisfaction rating exceeds 42? We can be at least 99% certain, but not as much as 99.9% certain.
WEEK 4 ASSIGNMENT 4
Chapter 9: 9.22 How do we decide whether to use a z test or a t test when testing a hypothesis about a population mean? By assuming the population is approximately normally distributed, we can use the z-test when the sample size is large (n >=30) or when we know the standard deviation sigma of the population, even if the sample

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