Question

A recent survey of 10 social networking sites has a mean of 12.67million visitors for a specific month.The standard deviation was 3.1million. Find the 95% confidence interval of the true mean. Assume the variable is normallydistributed. Round your answers to at least two decimal places.

Answer #1

Solution :

Given that,

t /2,df = 2.262

Margin of error = E = t/2,df * (s /n)

= 2.262 * (3.1 / 10)

Margin of error = E = 2.22

The 95% confidence interval estimate of the population mean is,

- E < < + E

12.67 - 2.22 < < 12.67 + 2.22

10.45 < < 14.89

**(10.45 , 14.89)**

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