Question

Estimate the population mean by finding a 99.9% confidence interval given a sample of size 70, with a mean of 60.5 and a standard deviation of 9.4.

Preliminary:

1. Is it safe to assume that n≤0.05 of all subjects in the population?
   • Yes
   • No
   
   
2. Is n≥30?
   • No
   • Yes

Confidence Interval: What is the 99.9% confidence interval to estimate the population mean? Enter your answer as an open-interval (i.e., parentheses) to one decimal place.

99.9% C.I. =

Answer 1

Yes, n <= 0.05 N For all subjects in the population.

n = 70 >= 30

df = n = 1 = 70 - 1 = 69

From T table,

t critical value at 0.001 significance level with 69 df = 3.437

99.9% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.001 /2, 70- 1 ) = 3.437
60.5 ± 3.437 * 9.4/√(70)
Lower Limit = 60.5 - 3.437 * 9.4/√(70)
Lower Limit = 56.6
Upper Limit = 60.5 + 3.737 * 9.4/√(70)
Upper Limit = 64.4
99.9% Confidence interval is ( 56.6 , 64.4  )