An article in the Archives of Internal Medicine reported that in a sample of 247 men, 73 had elevated total cholesterol levels (more than 200 milligrams per deciliter). In a sample of 231 women, 45 had elevated cholesterol levels. Can you conclude that the proportion of men with elevated cholesterol levels is greater than the proportion of women with elevated cholesterol levels? Let pv1 denote the proportion of men with elevated cholesterol levels and pv2 denote the proportion of women with elevated cholesterol levels. Use the a=0.05 level of significance and the P-value method.

Question

Hv0 : [?]
Hv1 : [?]

Is this a right/left or two-tailed test?

Answer 1

To determine whether the proportion of men with elevated cholesterol levels is significantly greater than the proportion of women, you can follow these steps:

Step 1: Define the Hypotheses

1. Null Hypothesis (H0): The proportion of men with elevated cholesterol levels is equal to or less than the proportion of women. That is: \[ H_0: p_v1 \leq p_v2 \]

2. Alternative Hypothesis (H1): The proportion of men with elevated cholesterol levels is greater than the proportion of women. That is: \[ H_1: p_v1 > p_v2 \]

Step 2: Identify the Type of Test

Since you are testing whether one proportion is greater than another, this is a one-tailed test (specifically, a right-tailed test).

Summary

• Null Hypothesis (H0): \( p_v1 \leq p_v2 \)
• Alternative Hypothesis (H1): \( p_v1 > p_v2 \)
• Type of Test: Right-tailed test

Step 3: Further Analysis

Next, you would perform the calculations needed to evaluate these hypotheses (calculating the proportions, standard error, test statistic, and then finding the p-value). If the p-value is less than the alpha level of 0.05, you would reject the null hypothesis in favor of the alternative hypothesis, concluding that there is statistically significant evidence that the proportion of men with elevated cholesterol is greater than that of women.

Answer 2

Find the P-value. Round the answer to four decimal places.

Answer 3

To calculate the P-value for this problem, we’ll first need to compute the sample proportions of men and women with elevated cholesterol levels, the pooled proportion, and then the test statistic. Here’s how to do it step by step:

Step 1: Calculate the Sample Proportions

1. Men: \[ n_1 = 247, \quad x_1 = 73 \quad (proportion , p_v1 = \frac{x_1}{n_1} = \frac{73}{247} \approx 0.2955) \]

2. Women: \[ n_2 = 231, \quad x_2 = 45 \quad (proportion , p_v2 = \frac{x_2}{n_2} = \frac{45}{231} \approx 0.1948) \]

Step 2: Calculate the Pooled Proportion

The pooled proportion (\(p\)) is calculated as follows: \[ p = \frac{x_1 + x_2}{n_1 + n_2} = \frac{73 + 45}{247 + 231} = \frac{118}{478} \approx 0.246 \]

Step 3: Standard Error Calculation

The standard error (SE) for the difference in proportions is given by: \[ SE = \sqrt{p(1 - p) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.246(1 - 0.246) \left( \frac{1}{247} + \frac{1}{231} \right)} \]

Calculating this: \[ SE = \sqrt{0.246 \cdot 0.754 \left( \frac{1}{247} + \frac{1}{231} \right)} \] \[ = \sqrt{0.246 \cdot 0.754 \left( 0.00404865 + 0.004329 \right)} \] \[ = \sqrt{0.246 \cdot 0.754 \cdot 0.00837765} \] \[ = \sqrt{0.0015151} \approx 0.0389 \]

Step 4: Calculate the Test Statistic

The test statistic (Z) is calculated by: \[ Z = \frac{p_v1 - p_v2}{SE} = \frac{0.2955 - 0.1948}{0.0389} \approx \frac{0.1007}{0.0389} \approx 2.585 \]

Step 5: Calculate the P-value

Now, using the Z-score, we can find the P-value. Since this is a one-tailed test, we will find the probability that Z is greater than 2.585 using the standard normal distribution table or a calculator.

Using a standard normal distribution calculator, we find: \[ P(Z > 2.585) \approx 0.0048 \]

Conclusion

Thus, the P-value is approximately: \[ \boxed{0.0048} \] Rounded to four decimal places, it's \(0.0048\).

Answer 4

Do we reject Hv0

Answer 5

To determine whether to reject the null hypothesis \(H_0\) based on the P-value, you compare the P-value to your chosen significance level (\(\alpha\)). In this case, you set \(\alpha = 0.05\).

Decision Rule

• If the P-value is less than \(\alpha\), you reject the null hypothesis \(H_0\).
• If the P-value is greater than or equal to \(\alpha\), you fail to reject the null hypothesis \(H_0\).

Conclusion Based on the P-value

In your case:

• Calculated P-value: \(0.0048\)
• Significance level (\(\alpha\)): \(0.05\)

Since \(0.0048 < 0.05\), you reject the null hypothesis \(H_0\).

Final Conclusion

There is sufficient statistical evidence to conclude that the proportion of men with elevated cholesterol levels is greater than the proportion of women with elevated cholesterol levels at the \(0.05\) significance level.