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.)

For %=99

z-score =2.575

So, the width of the confidence interval=z*SE = 2.575*0.33=0.84975
Rest part as followed by the upper one.

  Edit