Question

Test the claim that the proportion of people who own cats is larger than 70% at the 0.10 significance level.

The null and alternative hypothesis would be:


H0:μ=0.7H0:μ=0.7

H1:μ≠0.7H1:μ≠0.7


H0:p≥0.7H0:p≥0.7

H1:p<0.7H1:p<0.7


H0:μ≥0.7H0:μ≥0.7

H1:μ<0.7H1:μ<0.7


H0:μ≤0.7H0:μ≤0.7

H1:μ>0.7H1:μ>0.7


H0:p≤0.7H0:p≤0.7

H1:p>0.7H1:p>0.7


H0:p=0.7H0:p=0.7

H1:p≠0.7H1:p≠0.7


The test is:


left-tailed


right-tailed


two-tailed


Based on a sample of 400 people, 75% owned cats


The test statistic is: (to 2 decimals)


The p-value is: (to 2 decimals)


Based on this we:


Reject the null hypothesis


Fail to reject the null hypothesis

Answer 1

H0:p≤0.7

H1:p>0.7

right tailed test

`z=\frac{0.75 -0.7}{\sqrt{(0.7(1-0.7))/(400)}}=2.182`

`p_v =P(z>2.182)=0.0146`

Since the p value is lower than the significance level given of 0.1 we have enough evidence to reject the null hypothesis.

Reject the null hypothesis

Explanation:

For this case we want to test if the the proportion of people who own cats is larger than 70% at the 0.10 significance level so then the best system of hypothesis are:

H0:p≤0.7

H1:p>0.7

And for this case if we analyze the alternative hypothesis we see that we are conducting a right tailed test

Data given

n=400 represent the random sample taken

`\hat p=0.75` estimated proportion for the people with cats

`p_o=0.7` is the value that we want to test

represent the significance level

z would represent the statistic

`p_v` represent the p value

The statistic is given by:

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

Replacing the data we got:

`z=\frac{0.75 -0.7}{\sqrt{(0.7(1-0.7))/(400)}}=2.182`

The p valye for this case would be:

`p_v =P(z>2.182)=0.0146`

Since the p value is lower than the significance level given of 0.1 we have enough evidence to reject the null hypothesis.

Reject the null hypothesis