Step 1: Check the Assumptions for a Z-Test of Two Population Proportions

For a z-test for two population proportions, the following conditions must hold:

1. Random Samples: The samples from both populations are independently and randomly selected.
2. Normality: The sample size should be large enough for the sampling distribution of the difference in proportions to be approximately normal. Specifically, we need:
   • $n_1 p_1 \geq 5$
   • $n_1 (1 - p_1) \geq 5$
   • $n_2 p_2 \geq 5$
   • $n_2 (1 - p_2) \geq 5$
     Where $n_1$ and $n_2$ are the sample sizes (100 each in this case).

Let's verify these conditions.

$n_1 = n_2 = 100, \quad p_1 = 0.40, \quad p_2 = 0.35$

• $n_1 p_1 = 100 \times 0.40 = 40 \geq 5$
• $n_1 (1 - p_1) = 100 \times (1 - 0.40) = 100 \times 0.60 = 60 \geq 5$
• $n_2 p_2 = 100 \times 0.35 = 35 \geq 5$
• $n_2 (1 - p_2) = 100 \times (1 - 0.35) = 100 \times 0.65 = 65 \geq 5$

Since all the conditions hold, the assumptions are met.

---

Step 2: Calculate the Z-Statistic

We are testing:

$H_0: P_1 = P_2 \quad \text{vs} \quad H_1: P_1 > P_2$

The test statistic for comparing two proportions is given by:

$z = \frac{(p_1 - p_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

where:

$\hat{p} = \frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac{100 \times 0.40 + 100 \times 0.35}{200} = \frac{40 + 35}{200} = 0.375$

Now, substitute into the formula:

$z = \frac{0.40 - 0.35}{\sqrt{0.375 \times (1 - 0.375) \left(\frac{1}{100} + \frac{1}{100}\right)}}$

$z = \frac{0.05}{\sqrt{0.375 \times 0.625 \times 0.02}}$

$z = \frac{0.05}{\sqrt{0.0046875}} = \frac{0.05}{0.06846} \approx 0.73$

So, the z-statistic is approximately:

$z = 0.73$

---

Step 3: Calculate the Critical Values

Since this is a one-tailed test (we are testing if $P_1 > P_2$), we need the critical z-values for two different significance levels:

• At $\alpha = 0.01$: Find the critical value $z_{\alpha = 0.01}$ for a one-tailed test.

  • From z-tables, $z_{0.01} \approx 2.33$.

• At $\alpha = 0.10$: Find the critical value $z_{\alpha = 0.10}$ for a one-tailed test.

  • From z-tables, $z_{0.10} \approx 1.28$.

---

Step 4: Decision to Accept or Reject $H_0$

We compare the calculated z-statistic ($z = 0.73$) with the critical values at each significance level.

• At $\alpha = 0.01$:

  • Critical value: $z_{0.01} = 2.33$
  • Since $0.73 < 2.33$, we fail to reject $H_0$ at the 1% significance level.

• At $\alpha = 0.10$:

  • Critical value: $z_{0.10} = 1.28$
  • Since $0.73 < 1.28$, we fail to reject $H_0$ at the 10% significance level.

---

Conclusion

At both significance levels ($\alpha = 0.01$ and $\alpha = 0.10$), we fail to reject the null hypothesis $H_0: P_1 = P_2$. This suggests that there is not enough evidence to conclude that $P_1$ is greater than $P_2$.

---

Related Questions

1. How do we interpret the p-value in hypothesis testing?
2. How is the pooled proportion used in two-proportion z-tests?
3. What are the differences between one-tailed and two-tailed hypothesis tests?
4. How do Type I and Type II errors relate to significance levels?
5. What alternatives to the z-test exist for small sample sizes?

---

Tip: A smaller p-value provides stronger evidence against the null hypothesis, making the choice of significance level critical in decision-making.