Question

5. A telephone company claims that 20% of its customers have at least two telephone lines. The company selects a random sample of 500 customers and finds that 88 have two or more telephone lines. At a 0.05 level of significance does the data support the claim? State the null and alternative hypotheses.

Answer 1

`z=\frac{0.176 -0.2}{\sqrt{(0.2(1-0.2))/(500)}}=-1.342`

We are conducting a bilateral test so then the p value would be:

`p_v =2*P(z<-1.342)=0.180`

For this case since the p value is higher than the significance level we FAIL to reject the null hypothesis and we can conclude that true proportion of customers have at least two telephone lines is not diffeent from 0.2 or 20%

Explanation:

Information given

n=500 represent the random sample taken

X=88 represent the people with two or more telephone lines

`\hat p=(88)/(500)=0.176` estimated proportion of people with two or more lines

`p_o=0.2` is the value that we want to test

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value

System of hypothesis

We want to check if the true proportion of people in a company with at least two telephone lines is 0.2 or no .:

Null hypothesis:
`p=0.2`

Alternative hypothesis:
`p \\eq 0.2`

The statistic is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

Replacing the info given we got:

`z=\frac{0.176 -0.2}{\sqrt{(0.2(1-0.2))/(500)}}=-1.342`

We are conducting a bilateral test so then the p value would be:

`p_v =2*P(z<-1.342)=0.180`

For this case since the p value is higher than the significance level we FAIL to reject the null hypothesis and we can conclude that true proportion of customers have at least two telephone lines is not diffeent from 0.2 or 20%