Question

A sample of 64 students from a large university is taken. The average age in the sample was 22 years with a standard deviation of 6 years. Construct a 95% confidence interval for the average age of the population.

Answer 1

Answer on Question #51527 – Math – Statistics and Probability

Question

A sample of 64 students from a large university is taken. The average age in the sample was 22 years with a standard deviation of 6 years. Construct a 95% confidence interval for the average age of the population.

Solution

$$n = 64$$ $$\bar{X} = 22$$ $$\sigma = 6$$ $$\gamma = 95\% \Rightarrow \alpha = 5\% = 0.05$$ $$S^2 = \frac{n}{n-1} \sigma^2$$ $$S = \sqrt{\frac{n}{n-1}} \sigma = \sqrt{\frac{64}{63}} \cdot 6 = 6.05$$ $$P \left( \bar{X} - t_{1-\frac{\alpha}{2}, n-1} - \frac{S}{\sqrt{n}} \leq \mu \leq \bar{X} + t_{1-\frac{\alpha}{2}, n-1} - \frac{S}{\sqrt{n}} \right) = 1 - \alpha$$

where $t_{1-\frac{\alpha}{2}, n-1} - 1 - \frac{\alpha}{2}$ - level quantile of Student's distribution with $n-1$ degrees of freedom. $t_{1-\frac{0.05}{2}, 100-1} = 1.984$.

Then

$$t_{1-\frac{\alpha}{2}, n-1} - \frac{S}{\sqrt{n}} = 1.984 \cdot \frac{6.05}{\sqrt{64}} = 1.5, \text{ so}$$ $$22 - 1.5 \leq \mu \leq 22 + 1.5$$ $$20.5 \leq \mu \leq 23.5$$

is 95% confidence interval for the average age of the population.

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