4. An article reported that in a sample of 244 men, 73 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 232 women, 44 had elevated cholesterol levels. Can we conclude at a = 0.05 that the proportion of people with elevated cholesterol levels differs between men and women? Please show all 4 steps of the classical approach clearly and interpret your conclusion.

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**Problem 4:**

An article reported that in a sample of 244 men, 73 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 232 women, 44 had elevated cholesterol levels. Can we conclude at \( \alpha = 0.05 \) that the proportion of people with elevated cholesterol levels differs between men and women? Please show all 4 steps of the *classical approach* clearly and interpret your conclusion.

### Explanation of the Classical Approach:

1. **State the Hypotheses:**
   - Null Hypothesis (\( H_0 \)): The proportion of men with elevated cholesterol levels is equal to the proportion of women with elevated cholesterol levels.
   - Alternative Hypothesis (\( H_a \)): The proportion of men with elevated cholesterol levels is different from the proportion of women with elevated cholesterol levels.

2. **Set the Significance Level (\( \alpha \))**:
   - \(\alpha = 0.05\)

3. **Compute the Test Statistic:**
   - Calculate the sample proportions:
     - Proportion of men, \( \hat{p}_1 = \frac{73}{244} \)
     - Proportion of women, \( \hat{p}_2 = \frac{44}{232} \)

   - Use the following formula to calculate the test statistic for the difference between two proportions:
     \[
     z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}
     \]
     where \(\hat{p}\) is the pooled sample proportion calculated as:
     \[
     \hat{p} = \frac{x_1 + x_2}{n_1 + n_2}
     \]
     and \( \hat{p}(1 - \hat{p}) \) is the combined proportion of successes and failures.

4. **Make a Decision:**
   - Compare the computed z-value with the critical z-value from the standard normal distribution for a two-tailed test at \( \alpha = 0.05 \) (which is approximately \(\pm 1.96\)).
   - If the absolute value of the computed z-value is greater than
Transcribed Image Text:**Problem 4:** An article reported that in a sample of 244 men, 73 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 232 women, 44 had elevated cholesterol levels. Can we conclude at \( \alpha = 0.05 \) that the proportion of people with elevated cholesterol levels differs between men and women? Please show all 4 steps of the *classical approach* clearly and interpret your conclusion. ### Explanation of the Classical Approach: 1. **State the Hypotheses:** - Null Hypothesis (\( H_0 \)): The proportion of men with elevated cholesterol levels is equal to the proportion of women with elevated cholesterol levels. - Alternative Hypothesis (\( H_a \)): The proportion of men with elevated cholesterol levels is different from the proportion of women with elevated cholesterol levels. 2. **Set the Significance Level (\( \alpha \))**: - \(\alpha = 0.05\) 3. **Compute the Test Statistic:** - Calculate the sample proportions: - Proportion of men, \( \hat{p}_1 = \frac{73}{244} \) - Proportion of women, \( \hat{p}_2 = \frac{44}{232} \) - Use the following formula to calculate the test statistic for the difference between two proportions: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}} \] where \(\hat{p}\) is the pooled sample proportion calculated as: \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \] and \( \hat{p}(1 - \hat{p}) \) is the combined proportion of successes and failures. 4. **Make a Decision:** - Compare the computed z-value with the critical z-value from the standard normal distribution for a two-tailed test at \( \alpha = 0.05 \) (which is approximately \(\pm 1.96\)). - If the absolute value of the computed z-value is greater than
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