The problem asks to conduct a hypothesis test for comparing two population proportions. Specifically, we want to test the null hypothesis:

$H_0: p_1 = p_2$

against the alternative hypothesis:

$H_a: p_1 < p_2$

Step 1: Calculate the sample proportions

First, calculate the sample proportions for each population.

For the first population:

$\hat{p}_1 = \frac{x_1}{n_1} = \frac{64}{506} \approx 0.1265$

For the second population:

$\hat{p}_2 = \frac{x_2}{n_2} = \frac{113}{610} \approx 0.1852$

Step 2: Pooled proportion

Since the null hypothesis assumes $p_1 = p_2$, we calculate the pooled sample proportion $\hat{p}$, which is the overall proportion of successes from both samples combined:

$\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{64 + 113}{506 + 610} = \frac{177}{1116} \approx 0.1587$

Step 3: Standard error

The standard error of the difference in sample proportions is calculated using the pooled proportion:

$SE = \sqrt{\hat{p} \cdot (1 - \hat{p}) \cdot \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}$

Substitute the values:

$SE = \sqrt{0.1587 \cdot (1 - 0.1587) \cdot \left(\frac{1}{506} + \frac{1}{610}\right)}$

$SE = \sqrt{0.1587 \cdot 0.8413 \cdot \left(0.001976 + 0.001639\right)}$

$SE = \sqrt{0.1587 \cdot 0.8413 \cdot 0.003615} \approx \sqrt{0.0004818} \approx 0.02195$

Step 4: Test statistic

Now, compute the standardized test statistic $z$ using the formula:

$z = \frac{\hat{p}_1 - \hat{p}_2}{SE}$

Substitute the values:

$z = \frac{0.1265 - 0.1852}{0.02195} \approx \frac{-0.0587}{0.02195} \approx -2.674$

Conclusion:

The standardized test statistic is approximately:

$z \approx -2.674$

Would you like more details, or do you have any questions? Here are some follow-up questions to expand on this:

1. How would you calculate the p-value for this test statistic?
2. What is the critical value for a left-tailed test at $\alpha = 0.005$?
3. How does the continuity correction affect the test statistic?
4. How do you interpret the result of this test?
5. What assumptions are necessary for using the normal distribution as an approximation?

Tip: Always check the assumptions (sample size and normality) when using the normal approximation for binomial tests.