In: Statistics and Probability
From a normal population, a sample of 39 items is taken. The sample mean is 12 and the sample standard deviation is 2. Construct a 99% confidence interval for the population mean.
solution
Given that,
= 12
s =2
n = 39
Degrees of freedom = df = n - 1 =39- 1 = 38
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t /2 df = t0.005,38 = 2.712 ( using student t table)
Margin of error = E = t/2,df * (s /n)
= 2.712 * (2 / 39) = 0.87
The 99% confidence interval mean is,
- E < < + E
12 -0.87 < < 12+ 0.87
11.13 < < 12.87
(11.13 , 12.87)