Question

A sample of 1300 computer chips revealed that 43% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that more than 40% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.10

level to support the company's claim?

State the null and alternative hypotheses for the above scenario.

Answer 1

Solution:

Null hypothesis: H₀: The percentage of chips does not fail in the first 1000 hours of their use is 40%.

Alternative hypothesis: H_a: The percentage of chips does not fail in the first 1000 hours of their use is more than 40%.

H₀: p = 0.40 versus H_a: p > 0.40

This is an upper tailed test.

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 = 1300

p̂ = x/n = 0.43

p = 0.40

q = 1 - p = 0.60

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.43 – 0.40)/sqrt(0.40*0.60/1300)

Z = 2.2079

P-value = 0.0136

(by using z-table)

P-value < α = 0.10

So, we reject the null hypothesis

There is sufficient evidence to conclude that the percentage of chips does not fail in the first 1000 hours of their use is more than 40%.