In: Statistics and Probability
An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has a an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.
Solution :
Given that,
Point estimate = sample mean =
=
Population standard deviation =
=
Sample size = n =
At 96% confidence level the z is ,
= 1 - 96% = 1 - 0.96 = 0.04
/ 2 = 0.04 / 2 = 0.02
Z/2 = Z0.02= 2.05 ( Using z table )
Margin of error = E = Z/2
* (
/n)
= 2.05 * ( 40/
30)
E= 14.9711
At 96% confidence interval estimate of the population mean
is,
- E <
<
+ E
780 - 14.9711 <
< 780 + 14.9711
765.0289 <
< 794.9711
( 765.0289 ,794.9711)