In a random sample of 40 refrigerators the mean repair cost

In a random sample of 40 refrigerators, the mean repair cost was $114.00 and the standard deviation was $15.80. A 90% confidence interval for the population mean repair cost is (109.89,118.11). Change the sample size to n=80. Construct a 90% confidence interval for the population mean repai cost. Which confidence interval is wider. Explain

Solution

When size n=80
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=114
Standard deviation( sd )=15.8
Sample Size(n)=80
Confidence Interval = [ 114 ± Z a/2 ( 15.8/ Sqrt ( 80) ) ]
= [ 114 - 1.645 * (1.766) , 114 + 1.645 * (1.766) ]
= [ 111.094,116.906 ]

It is narrower