In a random sample of 60 refrigerators the mean repair cost

In a random sample of 60 refrigerators, the mean repair cost was $118.00 and the population standard deviation is $16.70

A 90% confidence interval for the population mean repair cost is (114.45,121.55) Change the sample size to

nequals=120. Construct a 90% confidence interval for the population mean repair cost. Which confidence interval is wider? Explain.

(Round to two decimal places as needed.)

Solution

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    118          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    16.7          
n = sample size =    120          
              
Thus,              
              
Lower bound =    115.4924298          
Upper bound =    120.5075702          
              
Thus, the confidence interval is              
              
(   115.4924298   ,   120.5075702   ) [ANSWER]

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As we can see, the confidence interval for the sample of 60 is wider than the confidence interval for the sample of 120.

This is becasue the standard error of the mean is less when the sample size increases, as

standard error = s / sqrt(n).