Question

Suppose Connie owns a pet store in a large community that has over 50,000 pet owners. She believes that the proportion of men who own cats is different from the proportion of women who own cats. Connie decides to test this claim by performing a two-sample z- test for two proportions. Her hypotheses are given by H0:pm=pwH1:pm≠pw where H0 is the null hypothesis, and H1 is the alternative hypothesis. The variables pw and pm represents the population proportion of men and women cat owners, respectively. She randomly samples 510 men and 420 women who are pet owners in the community and finds that 214 men own cats and 215 women own cats. Determine the pooled sample proportion, ^p . Provide your answer precise to three decimal places.

Answer 1

Answer:

$\hat p=\frac{X_{m}+X_{w}}{n_{m}+n_{w}}=\frac{214+215}{510+420}=0.461$

$z=\frac{0.420-0.512}{\sqrt{0.461(1-0.461)(\frac{1}{510}+\frac{1}{420})}}=-2.80$  

Since is a two tailed side test the p value would be:  

$p_v =2*P(Z$  

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the two proportions are significantly different.  

Step-by-step explanation:

Data given and notation  

$X_{m}=214$ represent the number of men with cats

$X_{w}=215$ represent the number of women with cats

$n_{m}=510$ sample for men

$n_{f}=420$ sample for women

$\hat p_{m}=\frac{214}{510}=0.420$ represent the proportion of men with cats

$\hat p_{f}=\frac{215}{420}=0.512$ represent the proportion of women with cats

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportions are different, the system of hypothesis would be:  

Null hypothesis:$p_{m} = p_{f}$  

Alternative hypothesis:$p_{m} \neq p_{f}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{m}-p_{f}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{m}}+\frac{1}{n_{f}})}}$   (1)

Where $\hat p=\frac{X_{m}+X_{w}}{n_{m}+n_{w}}=\frac{214+215}{510+420}=0.461$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.420-0.512}{\sqrt{0.461(1-0.461)(\frac{1}{510}+\frac{1}{420})}}=-2.80$  

Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

Since is a two tailed side test the p value would be:  

$p_v =2*P(Z$  

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the two proportions are significantly different.