in a random sample of 40 refrigerators the mean repair cost was $131 in the popu
ID: 3073783 • Letter: I
Question
in a random sample of 40 refrigerators the mean repair cost was $131 in the population standard deviation is $15.80 construct a 95% confidence interval for the population mean repair cost interpret the results.
Score: 0 of 1 pt 15 of 35 (35 complete)V X6.1.38 Question Help In a random sample of 40 refrigerators, the mean repair cost was $131.00 and the population standard deviation is $15.80. Construct a 95% confidence interval for the population mean repair cost. Interpret the results. Construct a 95% confidence interval for the population mean repair cost. The 95% confidence interval is ( L Tests (Round to two decimal places as needed.) ontents Success ia Library ons Enter your answer in the edit fields and then click Check Answer Check Answer Clear All part remaining Question B2 (1/1) VQuestion 31 (1/1) untion 29 (0.50/1 Question 30 (1/1)Explanation / Answer
Mean = 131
S.D. = 15.80
= M ± Z(sM)
where:
M = sample mean
Z = Z statistic determined by confidence level
sM = standard error = (s2/n)
M = 131
t = 1.96 @ 95% confidence interval
sM = (15.82/40) = 2.5
= M ± Z(sM)
= 131 ± 1.96*2.5
= 131 ± 4.9
CI [126.1, 135.9].
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