Question

A publisher reports that 45% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 370 found that 40% of the readers owned a laptop. Is there sufficient evidence at the 0.02 level to support the executive's claim?

Answer 1

At a significance level of 0.02, there is not enough evidence to support the claim that the percentage of readers that own a laptop is significantly different from 45%.

P-value = 0.06

Explanation:

This is a hypothesis test for a proportion.

The claim is that the percentage of readers that own a laptop is significantly different from 45%.

Then, the null and alternative hypothesis are:

`H_0: \pi=0.45\\\\H_a:\pi\\eq 0.45`

The significance level is 0.02.

The sample has a size n=370.

The sample proportion is p=0.4.

The standard error of the proportion is:

`\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.45*0.55)/(370)}\\\\\\ \sigma_p=√(0.000669)=0.026`

Then, we can calculate the z-statistic as:

`z=(p-\pi+0.5/n)/(\sigma_p)=(0.4-0.45+0.5/370)/(0.026)=(-0.049)/(0.026)=-1.881`

This test is a two-tailed test, so the P-value for this test is calculated as:

`\text{P-value}=2\cdot P(z<-1.881)=0.06`

As the P-value (0.06) is greater than the significance level (0.02), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.02, there is not enough evidence to support the claim that the percentage of readers that own a laptop is significantly different from 45%.