A random sample of 100 light bulbs has a mean lifetime of 3000 hours. Assume that the population standard deviation of the lifetime is 500 hours. Construct a 95% confidence interval estimate of the mean lifetime.

Solution

Given that sample size $n = 100$, sample mean $\overline{X}= 3000$ and population standard deviation $\sigma = 500$.

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

Step 2 Given information

Given that sample size $n =100$, sample mean $\overline{X}=3000$ and population standard deviation is $\sigma = 500$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$

where $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

Z-critical value
Z-critical value

Step 5 Compute the margin of error

The margin of error for mean is

$$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 1.96 \frac{500}{\sqrt{100}} \\ & = 98. \end{aligned} $$

Step 6 Determine the confidence interval

$95$ % confidence interval estimate for population mean is

$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 3000 - 98 & \leq \mu \leq 3000 + 98\\ 2902 & \leq \mu \leq 3098. \end{aligned} $$

Thus, $95$% confidence interval for the mean lifetime is $(2902,3098)$.

Further Reading