Assume that a sample is used to estimate a population mean μ . Find the 99.9% confidence interval for a sample of size 52 with a mean of 80.8 and a standard deviation of 17.2. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
99.9% C.I. =
Solution :
Given that,
Point estimate = sample mean = = 80.8
sample standard deviation = s = 17.2
sample size = n = 52
Degrees of freedom = df = n - 1 = 51
At 99.9% confidence level the t is ,
= 1 - 999% = 1 - 0.999 = 0.001
/ 2 = 0.001 / 2 = 0.0005
t /2,df = t0.0005,24 = 3.492
Margin of error = E = t/2,df * (s /n)
= 3.492* (17.2 / 52)
= 8.3
The 99% confidence interval estimate of the population mean is,
- E < < + E
80.8 - 8.3 < < 80.8 + 8.3
72.5 < < 89.1
99.9% C.I. = (72.5 , 89.1)
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