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In a sample of 246 men, 77 had high BP.

In a sample of 232 women, 41 had high BP
Can you conclude that the proportion of people with high BP differs between men and women? Use a=0.01
State the null and alternate hypothesis

3 answers

To determine if the proportion of people with high blood pressure (BP) differs between men and women, we can set up our hypotheses as follows:

Null Hypothesis (H0)

The proportion of men with high BP is equal to the proportion of women with high BP: \[ H_0: p_m = p_w \]

Alternative Hypothesis (H1)

The proportion of men with high BP is not equal to the proportion of women with high BP: \[ H_1: p_m \neq p_w \]

Where:

  • \( p_m \) is the proportion of men with high BP.
  • \( p_w \) is the proportion of women with high BP.

Sample Data

  • Men:

    • Sample size (\( n_m \)) = 246
    • Number with high BP = 77
    • Sample proportion (\( \hat{p}_m \)) = \( \frac{77}{246} \)

  • Women:

    • Sample size (\( n_w \)) = 232
    • Number with high BP = 41
    • Sample proportion (\( \hat{p}_w \)) = \( \frac{41}{232} \)

Significance Level

The significance level (\( \alpha \)) for this test is 0.01.

Next, we would typically conduct a hypothesis test (like a z-test for proportions) to evaluate these hypotheses using the sample proportions and their respective sample sizes.

Find the pvalue round to 4 places

To find the p-value for the hypothesis test comparing the proportions of men and women with high blood pressure, we will use the two-proportion z-test. Here are the steps to compute it:

Step 1: Compute Sample Proportions

  1. Proportion of men with high BP: \[ \hat{p}_m = \frac{77}{246} \approx 0.3122 \]

  2. Proportion of women with high BP: \[ \hat{p}_w = \frac{41}{232} \approx 0.1767 \]

Step 2: Calculate the Combined Sample Proportion

\[ \hat{p} = \frac{x_m + x_w}{n_m + n_w} = \frac{77 + 41}{246 + 232} = \frac{118}{478} \approx 0.246 \]

Step 3: Calculate the Standard Error (SE)

\[ SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \left( \frac{1}{n_m} + \frac{1}{n_w} \right)} \] Substituting the values: \[ SE = \sqrt{0.246 \cdot (1 - 0.246) \left( \frac{1}{246} + \frac{1}{232} \right)} \]

Calculating \( 1 - \hat{p} \): \[ 1 - 0.246 \approx 0.754 \]

Calculating \( \frac{1}{n_m} + \frac{1}{n_w} \): \[ \frac{1}{246} + \frac{1}{232} \approx 0.004065 + 0.004310 \approx 0.008375 \]

Then, substituting into the SE formula: \[ SE = \sqrt{0.246 \cdot 0.754 \cdot 0.008375} \approx \sqrt{0.00151} \approx 0.03886 \]

Step 4: Compute the Z-statistic

\[ z = \frac{\hat{p}_m - \hat{p}_w}{SE} = \frac{0.3122 - 0.1767}{0.03886} \approx \frac{0.1355}{0.03886} \approx 3.49 \]

Step 5: Find the p-value

For a two-tailed test, we look at \( P(Z > |3.49|) \).

Using a standard normal distribution table or calculator: At \( z \approx 3.49 \), the area in one tail is approximately 0.0002.

Since it's a two-tailed test: \[ p\text{-value} = 2 \times P(Z > 3.49) \approx 2 \times 0.0002 = 0.0004 \]

Conclusion

The p-value is approximately 0.0004 (rounded to four decimal places). Since this value is less than the significance level of \( 0.01 \), we would reject the null hypothesis, indicating a statistically significant difference between the proportions of men and women with high BP.