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A recent survey of 8 social networking sites has a mean of 13.1 million visitors for a specific month with a standard deviation of 4.1 million. Find the 95% confidence interval of the true mean and interpret the result.

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Answer: The formula for a confidence interval for a population mean, when the standard deviation is known, is:

CI = x ± z * (σ/√n)

where:

x = sample mean = 13.1 million

o = standard deviation = 4.1 million

n = sample size = 8

z = z-value for the desired confidence level. For a 95% confidence level, the z-value is approximately 1.96 (from standard normal distribution tables).

Now, let's plug in the values and calculate the confidence interval:

CI = 13.1 ± 1.96 * (4.1/√8)

CI = 13.1 ± 1.96 * (4.1/2.828)

CI = 13.1 ± 1.96 * 1.450

CI = 13.1 ± 2.84

So, the 95% confidence interval for the true mean is (13.1 - 2.84, 13.1 + 2.84) = (10.26, 15.94).

Interpretation:

We are 95% confident that the true mean number of visitors to the 8 social networking sites for a specific month is between 10.26 million and 15.94 million. This means that if we were to take many samples of the same size from this population, about 95% of the confidence intervals calculated from those samples would contain the true population mean.

User Lesnik
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