In: Statistics and Probability
A recent survey of 10 major news outlet websites has a mean of 14.2 million visitors for a specific month. The sample standard deviation was 3.9 million. Find the 95% confidence interval of the true mean.
Solution :
sample size = n = 10
Degrees of freedom = df = n - 1 = 9
t
/2,df = 2.262
Margin of error = E = t/2,df
* (s /
n)
= 2.262 * (3.9 /
10)
Margin of error = E = 2.8
The 95% confidence interval estimate of the population mean is,
- E <
<
+ E
14.2 - 2.8 <
< 14.2 + 2.8
11.4 <
< 17
(11.4, 17)