Question

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

Answer 1

The working equation is shown in the image. The z for a 95% confidence level is 1.96. The bar x is 13.1 million, sigma is 4.1 million and n is 8. Substituting,


$13.1 million - (1.96)( \frac{4.1 million}{ \sqrt{8} }) \ \ \textless \ true mean \ \textless$$13.1 million+ (1.96)( \frac{4.1 million}{ \sqrt{8} })$

$10.26 \ million \ \ \textless \ true \ mean \ \ \textless \ \ 15.94 \ million$

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Answer 2

The 95% confidence interval of the true mean for the considered case is : [9.672 × 10^6,  16.528 × 10^6]

How to calculate confidence interval for population mean for small sample?

If the sample size is given to be n < 30, then for finding the confidence interval for mean of population from this small sample, we use t-statistic.

• Let the sample mean given as $\overline{x}$ and
• The sample standard deviation s, and
• The sample size = n, and
• The level of significance = $\alpha$

Then, we get the confidence interval in between the limits

$\overline{x} \pm t_{\alpha/2}\times \dfrac{s}{\sqrt{n}}$

where $t_{\alpha/2}$ is the critical value of 't' that can be found online or from tabulated values of critical value for specific level of significance and degree of freedom n - 1.

For this case, we're specified that:

• The sample mean given as $\overline{x}$ = 13.1 millioni visitors
• The sample standard deviation s = 4.1 million visitors
• The sample size = n = 8

• The level of significance = $\alpha$ = 100 - 95% = 5% = 0.05 (converted percent to decimal).
  

Since the sample size is 8 < 30, so we will use t statistic for finding the confidence interval of the true mean (mean of the population).

The degree of freedom is n - 1 = 8 - 1 = 7

At d.f. 7 and level of significance 0.05, the critical value of t is obtained (from tables of critical value of t) as $t_{(0.05/2), 7} = 2.365$

Thus, we get:

$CI = \overline{x} \pm t_{\alpha/2}\times \dfrac{s}{\sqrt{n}}\\CI = 13.1 \times 10^6 \pm 2.365 \times \dfrac{4.1 \times 10^6}{\sqrt{8}}\\\\CI \approx [13.1 \times 10^6 - 3.428 \times 10^6, 13.1 \times 10^6 + 3.428 \times 10^6]\\\\CI = [9.672 \times 10^6, 16.528 \times 10^6]$

Thus, the 95% confidence interval of the true mean for the considered case is : [9.672 × 10^6,  16.528 × 10^6]

Learn more about confidence interval of population for small sample here:

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