Test Statistics

[Hasan]

A major swimming chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5000 of the 20000 households in the area have swimming pools. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have swimming pools. The value of the test statistic in this problem is approximately equal to

 

[Dason]

Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.

 

[Hasan]

Is there anyone who can solve this question?

 

[Hasan]

I have got the following.
H0=.25
H1>.25
Alfa =.05

z-value 2.8
one-tailed p-value 0.0026

95% confidence internvals
upper Inf
lower 0.2089

Reject the null hypothesis

Here, the value of test statistic is the p value which is 0.0026

Is this solution correct?

 

[GretaGarbo]

Is it possible to have a p-value of 0.0026 when the lower confidence limit is 0.2089 and H0=.25?

(I pasted in in Google: 96/300 -1.64*sqrt(0.32*0.68/300)
= 0.27 What does it mean?

And also; has there ever been an area where 5000 of the 20000 households in the area have swimming pools? Where one quarter of the households have a swimming pool?

(Ask your teacher about the realism of his homework.)