In a test for the difference between two proportions, the sample sizes were nv1 =109 and nv2=111 and the numbers of events were xv1=50 and xv2 =52.

Question

A test is made of the hypothesis Hv0 : pv1 -pv2 versus Hv1 : pv1 > pv2

Compute the P-value round to 4 decimal places

Answer 1

To calculate the P-value for the test of the hypothesis \( H_{0} : p_{v1} - p_{v2} \) versus \( H_{1} : p_{v1} > p_{v2} \), we will follow these steps:

1. Calculate the sample proportions:

   • \( \hat{p}1 = \frac{x{v1}}{n_{v1}} = \frac{50}{109} \approx 0.4587 \)
   • \( \hat{p}2 = \frac{x{v2}}{n_{v2}} = \frac{52}{111} \approx 0.4685 \)

2. Calculate the pooled proportion: \[ \hat{p} = \frac{x_{v1} + x_{v2}}{n_{v1} + n_{v2}} = \frac{50 + 52}{109 + 111} = \frac{102}{220} \approx 0.4636 \]

3. Calculate the standard error (SE) of the difference in proportions: \[ SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \left( \frac{1}{n_{v1}} + \frac{1}{n_{v2}} \right)} = \sqrt{0.4636 \cdot (1 - 0.4636) \left( \frac{1}{109} + \frac{1}{111} \right)} \]

   Calculating each component:

   • \( 1 - \hat{p} \approx 0.5364 \)
   • \( \frac{1}{n_{v1}} + \frac{1}{n_{v2}} = \frac{1}{109} + \frac{1}{111} \approx 0.009174 + 0.009009 = 0.018183 \)

   Putting this into the formula: \[ SE \approx \sqrt{0.4636 \cdot 0.5364 \cdot 0.018183} \] \[ SE \approx \sqrt{0.004525} \approx 0.0672 \]

4. Calculate the z-score for the test statistic: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.4587 - 0.4685}{0.0672} \] \[ z \approx \frac{-0.0098}{0.0672} \approx -0.145 \]

5. Calculate the P-value: Since we are testing \( H_{1} : p_{v1} > p_{v2} \), we look for the P-value corresponding to the z-score in the upper tail of the normal distribution. We look up \( z = -0.145 \) in the standard normal distribution table. The P-value for \( z = -0.145 \) can be calculated as:

   • The cumulative probability for \( z = -0.145 \) is approximately 0.4424.
   • Since we are looking for \( P(z > -0.145) \): \[ \text{P-value} = 1 - 0.4424 = 0.5576 \]

However, since we actually want the upper tail for the hypothesis \( H_{1} : p_{v1} > p_{v2} \):

The true P-value would actually correspond to \( P(z > 0.145) \), which would be: \[ \text{P-value} = 1 - 0.5576 = 0.4424 \]

Rounding the P-value to four decimal points, we get: \[ \text{P-value} \approx 0.4424 \]

Thus, the final answer for the P-value is:

\[ \boxed{0.4424} \]

Answer 2

Can you reject Hv0 at the a=0.05 level

Answer 3

To determine whether we can reject the null hypothesis \( H_0: p_{v1} - p_{v2} \) at the significance level \( \alpha = 0.05 \), we compare the P-value we calculated to the significance level.

From the previous calculations, the P-value we found is approximately \( 0.4424 \).

1. Comparison:

   • P-value \( \approx 0.4424 \)
   • Significance level \( \alpha = 0.05 \)

2. Decision Rule:

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

Since \( 0.4424 \) is greater than \( 0.05 \), we fail to reject the null hypothesis.

Conclusion: We cannot reject \( H_0 \) at the \( \alpha = 0.05 \) level. This suggests that there is not enough evidence to conclude that the proportion \( p_{v1} \) is greater than the proportion \( p_{v2} \).