Question

Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, and the sample mean was found to be 738.44 with a standard deviation of 38.20 and a standard error on the mean of 5.40. What is the p-value for this test (to 4 decimals)?

Answer 1

Solution :

Given that ,

$\mu$ = 750

= 738.44

T

$\sigma$ = 38.20

n = 50

The null and alternative hypothesis is ,

H₀ :  $\mu$ = 750

H_a : $\mu$ < 750

This is the left tailed test .

Test statistic = z

= (
T
- $\mu$ ) / $\sigma$ / $\sqrt$ n

= ( 738.44 - 750) / 38.20 / $\sqrt$ 50

= -2.14

The test statistic = -2.14

P - value = P (Z < -2.14 ) = 0.0162

The P-value for the test = 0.0162