In: Statistics and Probability
a recent survey of 10 networking sites has a mean of 10 million visitors for a specific month. The sample standard deviation was 2.2 million. Find the 90% confidence level
Solution :
Given that,
Point estimate = sample mean = = 10
sample standard deviation = s = 2.2
sample size = n = 10
Degrees of freedom = df = n - 1 = 10 -1 =9
At 90% confidence level
= 1-0.90% =1-0.9 =0.10
/2
=0.10/ 2= 0.05
t/2,df
= t0.05,9= 1.83
t /2,df = 1.83
Margin of error = E = t/2,df * (s /n)
= 1.83 * ( 2.2/ 10)
Margin of error = E = 1.28
The 90% confidence interval estimate of the population mean is,
- E < < + E
10 -1.28 < < 10+1.28
8.72 < < 11.28
(8.72,11.28)