Answer to Question #325518 in Statistics and Probability for wiwili

Question

"An electrical firm produces light bulbs that have a length of life that is approximately normally distributed with a mean of 680 hours and a population standard deviation of 26 hours. A new version of light bulbs is being produced and is assumed to be better than the previous version. To test this claim, a random sample of 93 new light bulbs are tested. Would you agree with this claim if the random sample showed an average of 920 hours? Use a 0.1 level of significance.

What are the given? Write only the number. :

population mean: Blank 1 hours

population standard deviation: Blank 2 hours

sample size: Blank 3

sample mean: Blank 4 hours

level of significance: Blank 5

What is the critical value?

z: Blank 6

What is the value of the calculated z? Round your answer to the nearest hundredths.

z: Blank 7"

Answer 1

population mean: 680 hours

population standard deviation: 26 hours

sample size 93

sample mean: 920 hours

level of significance 0.1

Null and alternative hypotheses:

$H_0: \mu\le680$

$H_1: \mu > 680$

Because σ is known and n=52>30,n=52>30, we can use the z-test.

z:$Z=\frac{X-\mu}{\sigma/\sqrt{n}}=\frac{920-680}{26/\sqrt{93}}=89$

In z-table, the area corresponding to z=89 is 1. Because the test is a right-tailed test, the P-value is equal to the area to the right of z=89, so, P=1-1=0

Because the P-value is less than level of significance =0.1, we reject the null hypothesis, there is enough evidence at the 10% level of significance to support the claim that the new version of light bulbs to be better than the previous version (the mean of length of life is more than 680 hours).