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Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .025 significance level.



The null and alternative hypothesis would be:

$\begin{array}{l}

H_0: p_M=p_F \quad H_0: p_M=p_F \quad H_0: \mu_M=\mu_F \quad H_0: \mu_M=\mu_F \quad H_0: \mu_M=\mu_F \quad H_0: p_M=p_F \\

H_1: p_Mp_F \quad H_1: \mu_M>\mu_F \quad H_1: \mu_M<\mu_F \quad H_1: \mu_M
eq \mu_F \quad H_1: p_M
eq p_F

\end{array}$



The test is:

left-tailed / right-tailed / two-tailed



Based on a sample of 60 men, 30% owned cats.

Based on a sample of 40 women, 40% 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

Verified Answer :

  • The hypotheses are \(H_0: p_M=p_F\) and \(H_1: p_M < p_F\), representing a left-tailed test.
  • The calculated test statistic is \(z \approx \boxed{-1.03}\).
  • The corresponding p-value is approximately \(\boxed{0.15}\).
  • Because the p-value is greater than the significance level \(\alpha = 0.025\), we \(\boxed{\text{Fail to reject the null hypothesis}}\).

Explanation

  1. Formulate Hypotheses and Identify Test Type

To test the claim that the proportion of men who own cats (\(p_M\)) is smaller than the proportion of women who own cats (\(p_F\)), we set up our hypotheses based on these proportions. The null hypothesis (\(H_0\)) represents the assumption of no difference, and the alternative hypothesis (\(H_1\)) represents the claim we are testing:

  • Null Hypothesis (\(H_0\)): \(p_M = p_F\)
  • Alternative Hypothesis (\(H_1\)): \(p_M < p_F\)

Since the alternative hypothesis specifies "smaller than," we are conducting a left-tailed test. The significance level for this test is \(\alpha = 0.025\).

  1. Calculate Test Statistic

We are given:

  • Men: \(n_M = 60\), \(\hat{p}_M = 0.30\), resulting in \(x_M = 60 \times 0.30 = 18\) cat owners.
  • Women: \(n_F = 40\), \(\hat{p}_F = 0.40\), resulting in \(x_F = 40 \times 0.40 = 16\) cat owners.

First, we calculate the pooled sample proportion \(\hat{p}\), which represents the combined proportion of cat owners across both groups:
\[\hat{p} = \frac{x_M + x_F}{n_M + n_F} = \frac{18 + 16}{60 + 40} = \frac{34}{100} = 0.34\]
Next, we calculate the standard error (\(SE\)) for the difference between two proportions:
\[SE = \sqrt{\hat{p}(1-\hat{p}) \left( \frac{1}{n_M} + \frac{1}{n_F} \right)} = \sqrt{0.34 \times 0.66 \times \left( \frac{1}{60} + \frac{1}{40} \right)} \approx 0.0967\]
Finally, the test statistic \(z\) is:
\[z = \frac{\hat{p}_M - \hat{p}_F}{SE} = \frac{0.30 - 0.40}{0.0967} \approx -1.03\]

  1. Determine P-value and Make a Decision

Using the standard normal distribution table for a left-tailed test with \(z \approx -1.03\), we find the p-value, which is the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true:
\[P(Z < -1.03) \approx 0.15\]
Comparing the p-value (\(0.15\)) to the significance level (\(\alpha = 0.025\)):
Since \(0.15 > 0.025\), we fail to reject the null hypothesis. There is not sufficient evidence at the \(0.025\) level to support the claim that the proportion of men who own cats is smaller than the proportion of women who own cats.

Examples

In market research, you might want to know if one demographic is more likely to buy a product than another. This hypothesis test helps determine if an observed difference in purchase rates is statistically significant or just a result of random sampling variation. For example, comparing the preference for a brand between two age groups can guide where a company spends its advertising budget.

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