What is the test statistic and p-value?

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 0.01 significance level. Use pooling. The null and alternative hypothesis would be:

Based on a sample of 60 men, 25% owned cats

Based on a sample of 80 women, 30% owned cats

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Answer 1

Hi Kaitlyn,

Pooled proportion estimator is computed by:

p-hat_pooled=(p-hat₁n₁ + p-hat₂n₂)/(n₁ + n₂)

We also need standard error:

SE=sqrt[(p-hat(1-p-hat)sqrt((1/n₁) + (1/n₂))]

Finally, our z-test statistic is:

z=p-hat₁ - p-hat₁ - 0/SE

We also need to confirm that both proposed proportions, when multiplied with sample size, exceed 10. We can tell just by looking at these that they do (0.25*60)=15, 0.30*80=24.

Now, let's break down those formulas since they are somewhat messy. First:

p-hat_pooled=(p-hat₁n₁ + p-hat₂n₂)/(n₁ + n₂)

p-hat₁=first sample proportion

p-hat₂=second sample proportion

n₁=first sample size

n₂=second sample size

For our problem:

p-hat_pooled=(p-hat₁n₁ + p-hat₂n₂)/(n₁ + n₂)

p-hat₁=0.25

p-hat₂=0.3

n₁=60

n₂=80

p-hat_pooled=[(0.25*60) + (0.30*80)]/(60+80)

p-hat_pooled=0.28

Now, for standard error:

SE=sqrt[(p-hat(1-p-hat)sqrt((1/n₁) + (1/n₂))]

p-hat=p-hat_pooled

n₁=first sample size

n₂=second sample size

For our problem:

p-hat=0.28

n₁=60

n₂=80

SE=sqrt(0.28*0.72) * sqrt((1/60) + (1/80))

SE=0.45*0.17

SE=0.0765

Finally, the test statistic:

z=[(p-hat₁ - p-hat₂) - 0]/SE

p-hat₁=0.25

p-hat₂=0.3

SE=0.0765

z=(0.25-0.3)/0.0765

z= -0.65

From z-table, P(z<-0.65)=0.2578

Given that we have a two-sided test:

P=2(0.2578)

p=0.52 (2 decimals)

No significant difference in population proportions of men and women who own cats. I hope this helps.