A recent survey of 7 social networking sites has a

Annette Sabin

Annette Sabin

Answered question

2021-12-20

A recent survey of 7 social networking sites has a mean of 14.69 million visitors for a specifie month. The standard deviation was 4.4 millon. Find the 95% confidence interval of the true mean. Assume the variable is normally distributed. Round your answers to two decimal places.

Answer & Explanation

Jeremy Merritt

Jeremy Merritt

Beginner2021-12-21Added 31 answers

n=7
x^=14.69
s=4.4
95% CI of mean
μ=x^±zsn
=μ=14.69±1.964.47
=14.69±3.25
=(11.44, 17.94)
Steve Hirano

Steve Hirano

Beginner2021-12-22Added 34 answers

Why z=1.96?
nick1337

nick1337

Expert2021-12-28Added 777 answers

Because at 95% CI, z=1.96

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