A recent survey of 10 major news outlet websites has a mean of 14.2 million visitors...

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.

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Expert Solution

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 / $\sqrt$ n)

= 2.262 * (3.9 / $\sqrt$ 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)

orchestra answered 1 year ago

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