Question

A sample of 1700 computer chips revealed that 43% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 44% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Is there enough evidence at the 0.05 level to support the manager's claim?

A. There is sufficient evidence to support the claim that the percentage of chips that do not fail is different from 44%

B. There is not sufficient evidence to support the claim that the percentage of chips that do not fail is different from 44%

Answer 1

H0: p = 0.44

Ha: p 0.44

Test statistics

z = - p / sqrt( p( 1 - p) / n)

= 0.43 - 0.44 / sqrt( 0.44 * 0.56 / 1700)

= -0.83

This is test statistics value.

p-value = 2 * P( Z < z)

= 2 * P( Z < -0.83)

= 2 * [ 1 - P( Z < 0.83) ]

= 2 * ( 1 - 0.7967 )

= 0.4066

Since p-value > 0.05 significance level, we do not have sufficient evidence to reject the null hypothesis.

We conclude at 0.05 level that,

There is not sufficient evidence to support the claim that the percentage of chips that do not fail is

different from 44%.