Question: The mean of the sample of 65 customer satisfaction ratings in Table 1.7 is x=42.95. if you let u denote the mean of all possible customer satisfaction ratings for the XYZ-Box video game system and assume that cr equals 2.64:
a. Calculate 95 percent and 99 percent confidence intervals for u.
b. Using the 95 percent confidence interval, can we be 95 percent confident that u is at least 42 (recall that a very satisfied customer gives a rating of at least 42)? Explain.
c. Using the 99 percent confidence interval, can we be 99 percent confident that u is at least 42? Explain.
d. Based on your answers to parts b and c, how convinced are you that the mean satisfaction rating is at least 42? Edit
Answer: n=65, x-bar=42.95, standard deviation σ =2.64, %=95
Standard error, SE= σ/√n =2.64/√65=0.33
z-score = 1.96
So, the width of the confidence interval=z*SE = 1.96*0.33=0.6468
Lower Limit=x-bar - width= 42.95-0.6468= 42.3032
Upper Limit=x-bar + width= 42.95+0.6468= 43.5968
So, the confidence intervals is [42.3032, 43.5968] (Ans.)
So, the width of the confidence interval=z*SE = 2.575*0.33=0.84975
Rest part as followed by the upper one.
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