Question

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.02 significance level. the null and alternative hypothesis would be: h 0 : p m = p f h 1 : p m ≠ p f h 0 : p m = p f h 1 : p m < p f h 0 : μ m = μ f h 1 : μ m < μ f h 0 : p m = p f h 1 : p m > p f h 0 : μ m = μ f h 1 : μ m > μ f h 0 : μ m = μ f h 1 : μ m ≠ μ f the test is: left-tailed right-tailed two-tailed based on a sample of 20 men, 40% owned cats based on a sample of 20 women, 60% 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

Answer:

We need to conduct a hypothesis in order to check if  that the proportion of men who own cats is significantly different than the proportion of women who own cats , the system of hypothesis would be:  

Null hypothesis:$p_{m} = p_{f}$  

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

And the best option would be:

h 0 : p m = p f h 1 : p m ≠ p f

$z=\frac{0.4-0.6}{\sqrt{(\frac{0.4*(1-0.4)}{20}+\frac{0.6*(1-0.6)}{20})}}=-1.29$  

$p_v =2*P(Z<-1.29)=0.20$  

So the p value is a very high value and using any significance level for example $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis

Step-by-step explanation:

Data given and notation  

$n_{m}=20$ sample of male selected

$n_{f}=20$ sample of female selected

$\hat p_{m}=0.4$ represent the proportion of men with cats

$p_{f}=0.6$ 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  that the proportion of men who own cats is significantly different than the proportion of women who own cats , the system of hypothesis would be:  

Null hypothesis:$p_{m} = p_{f}$  

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

And the best option would be:

h 0 : p m = p f h 1 : p m ≠ p f

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

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

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic

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

$z=\frac{0.4-0.6}{\sqrt{(\frac{0.4*(1-0.4)}{20}+\frac{0.6*(1-0.6)}{20})}}=-1.29$  

Statistical decision

The significance level provided is  $\alpha=0.02$, but we can calculate the p value for this test.  

Since is a two sided test the p value would be:  

$p_v =2*P(Z<-1.29)=0.20$  

So the p value is a very high value and using any significance level for example $\alpha=0.02$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis