Question

construct a confidence interval for $p_1 - p_2$ at the given level of confidence.
$x_1 = 383$, $n_1 = 537$, $x_2 = 404$, $n_2 = 574$, $95%$ confidence
the researchers are $square%$ confident the difference between the two population proportions, $p_1 - p_2$, is between $square$ and $square$
(use ascending order. type an integer or decimal rounded to three decimal places as needed)

Explanation:

Step1: Calculate sample proportions

The sample proportion formula is $\hat{p}=\frac{x}{n}$.
For the first sample: $\hat{p}_1=\frac{383}{537}\approx0.713$.
For the second sample: $\hat{p}_2=\frac{404}{574}\approx0.704$.

Step2: Find the critical value

For a 95% confidence level, the significance level $\alpha = 1 - 0.95=0.05$, and $\alpha/2=0.025$.
The critical value $z_{\alpha/2}=z_{0.025} = 1.96$ (from the standard normal distribution table).

Step3: Calculate the margin of error

The formula for the margin of error $E$ is $E = z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$.
Substitute the values:
$\hat{p}_1(1 - \hat{p}_1)=0.713\times(1 - 0.713)=0.713\times0.287 = 0.205631$.
$\hat{p}_2(1 - \hat{p}_2)=0.704\times(1 - 0.704)=0.704\times0.296 = 0.208384$.
$\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}=\frac{0.205631}{537}+\frac{0.208384}{574}\approx0.000383 + 0.000363=0.000746$.
$E = 1.96\times\sqrt{0.000746}\approx1.96\times0.0273\approx0.0535$.

Step4: Calculate the confidence interval

The confidence interval for $p_1 - p_2$ is $(\hat{p}_1-\hat{p}_2 - E,\hat{p}_1-\hat{p}_2 + E)$.
$\hat{p}_1-\hat{p}_2=0.713 - 0.704 = 0.009$.
The lower bound is $0.009-0.0535=- 0.0445\approx - 0.045$.
The upper bound is $0.009 + 0.0535=0.0625\approx0.063$.

Answer:

The researchers are $95\%$ confident the difference between the two population proportions, $p_1 - p_2$, is between $-0.045$ and $0.063$.