Solution :
Given that,
Point estimate = sample mean = = 5.1
sample standard deviation = s = 18.1
sample size = n = 47
Degrees of freedom = df = n - 1 = 47-1 = 46
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.9 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,46 = 1.679
Margin of error = E = t/2,df * (s /n)
= 1.679 * (18.1 / 47)
= 4.43
The 90% confidence interval estimate of the population mean is,
- E < < + E
5.1 - 4.43 < < 5.1 + 4.43
0.67 < < 9.53
(0.67,9.53)
The 90% confidence interval estimate of the population mean is
0.67 mg/dl < < 9.53 mg/dl
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