Questions: The numbers of successes and the sample sizes for independent simple random samples from two populations are provided for a right-tailed test. Use a 90% confidence interval. Complete parts (a) through (d). x1=23, n1=90, x2=22, n2=100, alpha=0.05 Click here to view a table of areas under the standard normal curve for negative values of z. Click here to view a table of areas under the standard normal curve for positive values of z. Which of the following is the correct conclusion for the hypothesis test? A. At the 5% significance level, reject H0; the data provide sufficient evidence to accept Ha. B. At the 5% significance level, do not reject H0; the data do not provide sufficient evidence to accept Ha. C. At the 5% significance level, do not reject H0; the data provide sufficient evidence to accept Ha. D. At the 5% significance level, reject H0; the data do not provide sufficient evidence to accept Ha. E. Using the two-proportions z-procedures is not appropriate.

Solution Steps

To compare the proportions of two independent samples, we first calculate the Z-statistic using the formula:

\[ z = \frac{p_1 - p_2}{\sigma} \]

where \( p_1 = \frac{x_1}{n_1} = \frac{23}{90} \approx 0.2556 \) and \( p_2 = \frac{x_2}{n_2} = \frac{22}{100} = 0.22 \). The pooled proportion \( p_{pool} \) is given by:

\[ p_{pool} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{23 + 22}{90 + 100} = \frac{45}{190} \approx 0.2368 \]

The standard error \( \sigma \) is calculated as:

\[ \sigma = \sqrt{p_{pool} \cdot (1 - p_{pool}) \cdot \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \approx 0.0618 \]

Thus, the Z-statistic is:

\[ z = \frac{0.2556 - 0.22}{0.0618} \approx 0.5756 \]

For a right-tailed test at the significance level \( \alpha = 0.05 \), the critical value corresponds to the Z-score that leaves 5% in the right tail of the standard normal distribution. This value is:

\[ Z_{critical} = \Phi^{-1}(1 - \alpha) = \Phi^{-1}(0.95) \approx 1.645 \]

The P-value is calculated as:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(0.5756) \approx 0.2824 \]

Thus, the P-value is:

\[ P \approx 0.7176 \]

We compare the P-value to the significance level \( \alpha \):

\[ P = 0.7176 > \alpha = 0.05 \]

Since the P-value is greater than the significance level, we do not reject the null hypothesis \( H_0 \).

Final Answer