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a recent survey of 6 social networking sites has a mean of 9.05 million visitors for a specific month. the standard deviation was 5 million. find the 98% confidence interval of the true mean. assume the variable is normally distributed. round your answers to at least two decimal places.

Respuesta :

The 98% confidence interval of the true mean is 43.02 million < true mean < 13.80

Given,

Number of social networking sites being surveyed = 6

Sample mean = 9.05 million visitors per month

Standard deviation = 5 million

We have to find the 98% confidence interval of the true mean;

Here,

Sample mean, x = 9.05 million

Standard deviation, σ = 5 million

Sample size, n = 6

z score of 98% is 2.326

Now,

9.05 million - 2.326 × (5 million / √6) < true mean < 9.05 million + 2.326 × (5 million / √6)

43.02 million < true mean < 13.80

Therefore,

The 98% confidence interval of the true mean is 43.02 million < true mean < 13.80

Learn more about true mean here;

https://brainly.com/question/2253744

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