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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

User Kroehre
by
8.3k points

1 Answer

5 votes

Answer:

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

User Jonathan Soifer
by
8.0k points
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