A simple random sample of size n = 20 is drawn from a population that is normally distributed. The sample mean is found to be x = 66 and the sample standard deviation is found to be s = 10. Construct a 90% confidence interval about the population mean.
Solution :
Given that,
Point estimate = sample mean = = 66
sample standard deviation = s = 10
sample size = n = 20
Degrees of freedom = df = n - 1 = 20 - 1 = 19
At 90% confidence level
= 1 - 90%
=1 - 0.90 =0.10
/2
= 0.05
t/2,df
= t0.05,19 = 1.729
Margin of error = E = t/2,df * (s /n)
= 1.729 * (10 / 20)
Margin of error = E =3.87
The 90% confidence interval estimate of the population mean is,
± E
= 66 ± 3.87
= ( 62.13, 69.87 )
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