Question

In: Statistics and Probability

a recent survey of 10 networking sites has a mean of 10 million visitors for a...

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

Solutions

Expert Solution

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)


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