Question

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .01 significance level.

The null and alternative hypothesis would be:

H0:pM=pFH0:pM=pF
H1:pM<pFH1:pM<pF

The test is: left-tailed

Based on a sample of 20 men, 45% owned cats
Based on a sample of 60 women, 70% owned cats

The test statistic is: ___ (to 2 decimals)

The p-value is: ____ (to 2 decimals)

Answer 1

Let's write the given information:

n₁ = sample size of men = 20

n₂ =  sample size of women = 60

x₁ = n₁ * p₁ = number of men who owned cats = 20*0.45 = 9

x₂ = n₂ * p₂ = number of men who owned cats = 60*0.70 = 42

Let's used minitab:

Step 1: Click on Stat >>> Basic Statistics >>>2 Proportions...

Step 2: Select Summarized data

Fill the given information

Look the following picture ...

Then click on Option:

Look the following image:

Then click on OK again click on Ok

So we get the following output

From the above output

z = -2.01 ,

and p-value = 0.022

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.022 > 0.01 so we used 2nd rule.

That is we fail to reject null hypothesis

Conclusion: At 5% level of significance there are not sufficient evidence to conclude that the proportion of men who own cats is smaller than the proportion of women who own cats.