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

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

[tex]\hat p=0.75[/tex] estimated proportion for the people with cats

[tex]p_o=0.7[/tex] is the value that we want to test

represent the significance level

z would represent the statistic

[tex]p_v[/tex] represent the p value

The statistic is given by:

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

Replacing the data we got:

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

The p valye for this case would be:

[tex]p_v =P(z>2.182)=0.0146[/tex]

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