Solution :
Given that,
Point estimate = sample mean = = 68
sample standard deviation = s = 18
sample size = n = 21
Degrees of freedom = df = n - 1 = 21-1= 20
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.9 = 0.1
/ 2 = 0.1 / 2 = 0.05
t /2,df = t0.05,20 = 1.725
Margin of error = E = t/2,df * (s /n)
= 1.725 * (18 / 21)
= 6.78
The 90% confidence interval estimate of the population mean is,
- E < < + E
68 - 6.78 < < 68 + 6.78
61.22 < < 74.78
(61.22,74.78)
The lower bound is 61.22
The upper bound is 74.78
A simple random sample of size n=21 is drawn from a population that is normally distributed....
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