A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, is found to be 110, and the sample standard deviation S is found to be 25. Construct a 99% confidence interval for if the sample size n = 21
c )solution
Given that,
= 110
s =25
n =21
Degrees of freedom = df = n - 1 = 21- 1 =20
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2,df= t0.005, 20= 2.845 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.845 * ( 25/ 21) = 15.5207
The 99% confidence interval estimate of the population mean is,
- E < < + E
110 -15.5207 < < 110+ 15.5207
94.4793 < < 125.1507
( 94.4793,125.1507)
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