Answer to Question #241082 in Statistics and Probability for Darv

Question

A manufacturer of flashlight batteries claim that the average life of his product will exceed 40 hours. A company is willing to buy a very large shipment of batteries if the claim is true. A random sample of 36 batteries is tested, and it is found that the sample mean is 45 hours. If the population of the batteries has a standard deviation of 5 hours, is it likely that the batteries will be bought?

Answer 1

$H_0: \mu ≤ 40 \\ H_1: \mu > 40 \\ \bar{x} = 45 \\ \sigma= 5 \\ n = 36$

Right-tailed test.

Test-statistic:

$Z = \frac{\bar{x} -\mu}{\sigma / \sqrt{n}} \\ Z = \frac{45-40}{5 / \sqrt{36}} = 6.00$

Let use 0.01 level of significance.

$P-value=P(Z>6.0) \\ = 1 -P(Z<6.0) \\ = 1-0.999968 \\ = 0.000032$

Let us use α=0.01

Reject H₀ if p-value < α.

P-value = 0.000032 < α = 0.01

We reject H₀.

We can conclude that the average life of his product will exceed 40 hours.

A company will buy a very large shipment of batteries.