Male and female students were asked, "if you received $100 cash by mail, that was
addressed to your
neighbor but mistakenly delivered to you, would you return it to
your neighbor?".
Of 69 males sampled, 53 said yes, and of 131 females sampled, 120
said yes.
Test the claim that the two proportions are different at a = 0.05 level of
significance.

Question

21-05-2024
Mathematics

Answered

Male and female students were asked, "if you received $100 cash by mail, that was
addressed to your
neighbor but mistakenly delivered to you, would you return it to
your neighbor?".
Of 69 males sampled, 53 said yes, and of 131 females sampled, 120
said yes.
Test the claim that the two proportions are different at a = 0.05 level of
significance.

Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2024-05-21T23:50:37-04:00

To test the claim that the proportions of males and females who would return the [tex]$100 to their neighbor are different, we can perform a two-proportion z-test. Here is a step-by-step solution: ### Step 1: State the Hypotheses - Null Hypothesis ($H_0$): The proportions of males and females who would return the money are the same, i.e., $ p_1 = p_2 $. - Alternative Hypothesis ($H_A$): The proportions of males and females who would return the money are different, i.e., $ p_1 \neq p_2 $. ### Step 2: Calculate the Sample Proportions Let's calculate the sample proportions for males ($ \hat{p}_1 $) and females ($ \hat{p}_2 $). - For males: \[ \hat{p}_1 = \frac{53}{69} \approx 0.7681 \] - For females: \[ \hat{p}_2 = \frac{120}{131} \approx 0.9160 \] ### Step 3: Calculate the Pooled Sample Proportion The pooled sample proportion ($ \hat{p} $) is calculated by combining the successes and trials from both samples. \[ \hat{p} = \frac{53 + 120}{69 + 131} = \frac{173}{200} = 0.865 \] ### Step 4: Calculate the Standard Error The standard error (SE) for the difference between two proportions is given by: \[ SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \] where $ n_1 = 69 $ (number of males sampled) and $ n_2 = 131 $ (number of females sampled). \[ SE = \sqrt{0.865 \times 0.135 \left( \frac{1}{69} + \frac{1}{131} \right)} \] \[ SE = \sqrt{0.865 \times 0.135 \left( 0.0145 + 0.0076 \right)} \] \[ SE = \sqrt{0.865 \times 0.135 \times 0.0221} \approx \sqrt{0.002584} \approx 0.0508 \] ### Step 5: Calculate the Z-Score The z-score for the difference in proportions is calculated as follows: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \] \[ z = \frac{0.7681 - 0.9160}{0.0508} \approx \frac{-0.148}{0.0508} \approx -2.91 \] ### Step 6: Find the P-Value Since this is a two-tailed test, we need to find the p-value associated with the z-score using the standard normal distribution. The p-value can be calculated as: \[ p\text{-value} = 2 \times P(Z > |z|) \] From standard normal distribution tables or using a calculator, the area to the right of $ |z| = 2.91 $ is approximately 0.0018. \[ p\text{-value} = 2 \times 0.0018 = 0.0036 \] ### Step 7: Compare the P-Value with the Significance Level We compare the p-value to the significance level $ \alpha = 0.05 $. \[ 0.0036 < 0.05 \] ### Step 8: Conclusion Since the p-value is less than the significance level, we reject the null hypothesis. This means there is sufficient evidence to support the claim that the proportions of males and females who would return the money are different. Therefore, we conclude that the proportions of males and females who would return the $[/tex]100 to their neighbor are significantly different at the [tex]$ \alpha = 0.05 $[/tex] level of significance.