Question

The average monthly electric bill of a random sample of 100 residents of a city is $90 with a standard deviation of $24. Construct a 95% confidence interval for the mean monthly electric bills of all residents.

Answer 1

Answer on Question #51526 – Math – Statistics and Probability

Question

The average monthly electric bill of a random sample of 100 residents of a city is $90 with a standard deviation of $24. Construct a 95% confidence interval for the mean monthly electric bills of all residents.

Solution

$$n = 100$$ $$\bar{X} = 90$$ $$\sigma = 24$$ $$\gamma = 95\% \Rightarrow \alpha = 5\% = 0.05$$ $$S^2 = \frac{n}{n - 1} \sigma^2$$ $$S = \sqrt{\frac{n}{n - 1}} \sigma = \sqrt{\frac{100}{99}} 24 = 24.12$$ $$P \left( \bar{X} - t_{1 - \frac{\alpha}{2}, n - 1} \frac{S}{\sqrt{n}} \leq \mu \leq \bar{X} + t_{1 - \frac{\alpha}{2}, n - 1} \frac{S}{\sqrt{n}} \right) = 1 - \alpha$$

where $t_{1 - \frac{\alpha}{2}, n - 1} - 1 - \frac{\alpha}{2}$ -level quantile of Student's t-distribution with $n - 1$ degrees of freedom. $t_{1 - \frac{0.05}{2}, 100 - 1} = 1.984$. Then

$$t_{1 - \frac{\alpha}{2}, n - 1} \frac{S}{\sqrt{n}} = 1.984 \frac{24.12}{\sqrt{100}} = 4.785408$$ $$90 - 4.785408 \leq \mu \leq 90 + 4.785408$$ $$85.215 \leq \mu \leq 94.785$$

is 95% confidence interval for the mean monthly electric bills of all residents.

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