 A recent survey of 7 social networking sites has a Annette Sabin 2021-12-20 Answered
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.

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$$\displaystyle{n}={7}$$
$$\displaystyle\hat{{{x}}}={14.69}$$
$$\displaystyle{s}={4.4}$$
$$\displaystyle{95}\%$$ CI of mean
$$\displaystyle\mu=\hat{{{x}}}\pm{\frac{{{z}\cdot{s}}}{{\sqrt{{n}}}}}$$
$$\displaystyle=\mu={14.69}\pm{\frac{{{1.96}\cdot{4.4}}}{{\sqrt{{7}}}}}$$
$$\displaystyle={14.69}\pm{3.25}$$
$$\displaystyle={\left({11.44},\ {17.94}\right)}$$
Not exactly what you’re looking for? Steve Hirano
Why z=1.96? nick1337
Because at 95% CI, z=1.96