A simple random sample of size n=23 is drawn from a population that is normally distributed. The sample mean is found to be x bar=68 and the sample standard deviation is found to be s=15. Construct a 90?% confidence interval about the population mean.
The 90?% confidence interval is ?(nothing?,nothing?).
Solution :
Given that,
= 68
s = 15
n = 23
Degrees of freedom = df = n - 1 = 23 - 1 = 22
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
t /2,df = t0.05,22 = 1.717
Margin of error = E = t/2,df * (s /n)
= 1.717 * (15 / 23)
= 5.37
The 90% confidence interval estimate of the population mean is,
- E < < + E
68 - 5.37 < < 68 + 5.37
62.63 < < 73.37
(62.63 , 73.37)
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