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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