Question

A random sample of 10 mobile phone batteries has a lifetime with variance 16 months. Assuming the lifetime of batteries to be normally distributed, construct a 95% confidence interval for the variance of all such mobile phone batteries.

Answer 1

Answer on Question #41937, Math, Statistics and Probability

A random sample of 10 mobile phone batteries has a lifetime with variance 16 months. Assuming the lifetime of batteries to be normally distributed, construct a 95% confidence interval for the variance of all such mobile phone batteries.

Solution

For a confidence level $1 - \alpha$ confidence interval for the variance is

$$\frac{(n - 1)s^2}{\chi_{\frac{\alpha}{2}}^2} \leq \sigma^2 \leq \frac{(n - 1)s^2}{\chi_{1 - \frac{\alpha}{2}}^2},$$

where $s^2$ is a sample variance, $n$ is a sample size.

For a sample size of $n = 10$, we will have $df = n - 1 = 9$ degrees of freedom. For a $95\%$ confidence interval, we have $\alpha = 0.05$, which gives $2.5\%$ of the area at each end of the chi-square distribution. We find values of $\chi_{0.975}^2 = 2.700$ and $\chi_{0.025}^2 = 19.023$. This leads to the inequality for the variance

$$\frac{9 \cdot 16}{19.023} \leq \sigma^2 \leq \frac{9 \cdot 16}{2.700} \rightarrow 7.570 \leq \sigma^2 \leq 53.333.$$

Answer: $7.570 \leq \sigma^2 \leq 53.333$.

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