Question

A sample of 1100 computer chips revealed that 58% of the chips fail in the first 1000 hours of their use. The company's promotional literature claimed that less than 61% fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.01 level to support the company's claim?

State the null and alternative hypotheses for the above scenario.

Answer 1

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: About 61% fail in the first 1000 hours of their use.

Alternative hypothesis: Ha: Less than 61% fail in the first 1000 hours of their use.

H0: p = 0.61 versus Ha: p < 0.61

This is a lower tailed test.

We are given

Level of significance = α = 0.01

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

n = sample size = 1100

p̂ = x/n = 0.58

p = 0.61

q = 1 - p = 0.39

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.58 - 0.61)/sqrt(0.61*0.39/1100)

Z = -2.0400

Test statistic = -2.0400

P-value = 0.0207

(by using z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is not sufficient evidence to conclude that Less than 61% fail in the first 1000 hours of their use.