A simple random sample of size
n equals n=40
is drawn from a population. The sample mean is found to be
x overbar equals x=121.2
and the sample standard deviation is found to be
s equals s=12.4.
Construct a 99% confidence interval for the population mean.
Find the lower and upper bounds.
Solution :
Given that,
Point estimate = sample mean = = 121.2
sample standard deviation = s = 12.4
sample size = n = 40
Degrees of freedom = df = n - 1 = 39
t /2,df = 2.708
Margin of error = E = t/2,df * (s /n)
= 2.708 * (12.4 / 40)
Margin of error = E = 5.3
The 99% confidence interval estimate of the population mean is,
- E < < + E
121.2 - 5.3 < < 121.2 + 5.3
115.9 < < 126.5
(115.9 , 126.5)
Lower bound = 115.9
Upper bound = 126.5
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