Certainly! Let's walk through the process to decide whether the normal sampling distribution can be used and to test the claim about the difference between the two population proportions [tex]\( p_1 \)[/tex] and [tex]\( p_2 \)[/tex] at the significance level [tex]\( \alpha = 0.05 \)[/tex] using the given sample statistics.
### Step 1: Verify Conditions for Normal Approximation
To use the normal distribution, we need to ensure the sample sizes are large enough. Specifically, both [tex]\( np \)[/tex] and [tex]\( n(1-p) \)[/tex] should be greater than 5 for both samples. This is generally assessed using the sample proportions [tex]\( \hat{p}_1 \)[/tex] and [tex]\( \hat{p}_2 \)[/tex].
1. Calculate the sample proportions:
[tex]\[
\hat{p}_1 = \frac{x_1}{n_1} = \frac{65}{141} \approx 0.461
\][/tex]
[tex]\[
\hat{p}_2 = \frac{x_2}{n_2} = \frac{43}{176} \approx 0.244
\][/tex]
2. Check the conditions:
[tex]\[
n_1 \hat{p}_1 = 141 \times 0.461 \approx 65
\][/tex]
[tex]\[
n_1 (1 - \hat{p}_1) = 141 \times (1 - 0.461) \approx 76
\][/tex]
[tex]\[
n_2 \hat{p}_2 = 176 \times 0.244 \approx 43
\][/tex]
[tex]\[
n_2 (1 - \hat{p}_2) = 176 \times (1 - 0.244) \approx 133
\][/tex]
Since all values are greater than 5, the normal approximation is appropriate.
### Step 2: State the Hypotheses
- Null hypothesis [tex]\( H_0 \)[/tex]: [tex]\( p_1 = p_2 \)[/tex]
- Alternative hypothesis [tex]\( H_A \)[/tex]: [tex]\( p_1 \neq p_2 \)[/tex]
### Step 3: Calculate the Pooled Proportion
[tex]\[
\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{65 + 43}{141 + 176} = \frac{108}{317} \approx 0.341
\][/tex]
### Step 4: Calculate the Standard Error
[tex]\[
SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}
= \sqrt{0.341 (1 - 0.341) \left( \frac{1}{141} + \frac{1}{176} \right)}
\approx 0.054
\][/tex]
### Step 5: Calculate the Test Statistic (z-score)
[tex]\[
z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.461 - 0.244}{0.054} \approx 4.045
\][/tex]
### Step 6: Calculate the p-value
[tex]\[
\text{p-value} = 2 \times (1 - \text{CDF}(|z|))
\][/tex]
For [tex]\( z = 4.045 \)[/tex], the cumulative distribution function (CDF) value for a standard normal distribution is very close to 1.
[tex]\[
\text{p-value} \approx 2 \times (1 - 0.99997) = 2 \times 0.00003 = 0.00006
\][/tex]
### Step 7: Make a Decision
Compare the p-value to the significance level [tex]\( \alpha = 0.05 \)[/tex].
[tex]\[
0.00006 < 0.05
\][/tex]
Since the p-value is less than [tex]\( \alpha \)[/tex], we reject the null hypothesis.
### Conclusion
At the [tex]\( 0.05 \)[/tex] significance level, there is sufficient evidence to conclude that the difference between the two population proportions [tex]\( p_1 \)[/tex] and [tex]\( p_2 \)[/tex] is statistically significant. Therefore, we reject the claim that [tex]\( p_1 = p_2 \)[/tex].
