96% confidence interval

[logistic_guy]

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of $\displaystyle 40$ hours. If a sample of $\displaystyle 30$ bulbs has an average life of $\displaystyle 780$ hours, find a $\displaystyle 96\%$ confidence interval for the population mean of all bulbs produced by this firm.

 

[khansaheb]

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of $\displaystyle 40$ hours. If a sample of $\displaystyle 30$ bulbs has an average life of $\displaystyle 780$ hours, find a $\displaystyle 96\%$ confidence interval for the population mean of all bulbs produced by this firm.

show us your effort/s to solve this problem.

 

[logistic_guy]

Theorem.

The confidence interval is given by:

$\displaystyle \left(\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \ \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)$

where $\displaystyle z_{\alpha/2}$ is the $\displaystyle z$-value leaving an area of $\displaystyle \alpha/2$ to the right.

We are given:

$\displaystyle 1 - \alpha = 0.96$
$\displaystyle \alpha = 1 - 0.96 = 0.04$

Then,

$\displaystyle \frac{\alpha}{2} = \frac{0.04}{2} = 0.02$

$\displaystyle z_{0.02}$ (leaving an area of $\displaystyle 0.02$ to the right) = $\displaystyle z_{0.98}$.

Looking at the normal table, we find that $\displaystyle z_{0.98} = 2.05$.

Then, the $\displaystyle 96\%$ confidence interval is:

$\displaystyle \left(\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}}, \ \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)$

$\displaystyle \left(780 - 2.05\frac{40}{\sqrt{30}}, \ 780 + 2.05\frac{40}{\sqrt{30}}\right)$

$\displaystyle \left(765, \ 795\right)$