Respuesta :
Answer:
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
Step-by-step 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:
[tex]H_0: \pi=0.45\\\\H_a:\pi\neq 0.45[/tex]
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:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.45*0.55}{370}}\\\\\\ \sigma_p=\sqrt{0.000669}=0.026[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.4-0.45+0.5/370}{0.026}=\dfrac{-0.049}{0.026}=-1.881[/tex]
This test is a two-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=2\cdot P(z<-1.881)=0.06[/tex]
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%.
