Question

11.1.15 Question Help Construct a confidence interval for P1-P2 at the given level of confidence. X1 = 30, n = 269, X2 = 32, n = 300, 99% confidence and The researchers are % confident the difference between the two population proportions, P1 - P2, is between (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.)

Answer 1

The formula for calculating $(1-\alpha)100\%$ for the true population proportion difference is given by $(\hat{p}_{1}-\hat{p}_{2})\pm Z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{2})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}$

where,

sample proportion for sample 1, $\hat{p}_{1}=\frac{x_{1}}{n_{1}}=\frac{30}{269}=0.1115241636\approx \mathbf{0.11152}$

sample proportion for sample 2, $\hat{p}_{2}=\frac{x_{2}}{n_{2}}=\frac{32}{300}=0.1066666667\approx \mathbf{0.10667}$

Critical value: For 99% confidence, $\alpha=1-0.99=0.01$

$Z_{\frac{\alpha}{2}}=Z_{\frac{0.01}{2}}=2.575829\approx \mathbf{2.576}$

Calculation for 99% confidence interval for $\hat{p}_{1}-\hat{p}_{2}$ :

$\Rightarrow (\hat{p}_{1}-\hat{p}_{2})\pm Z_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{2})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}$

$\Rightarrow \left(0.11152-0.10667\right)\pm2.576\sqrt{\frac{0.11152\left(1-0.11152\right)}{269}+\frac{0.10667\left(1-0.10667\right)}{300}}$

$\Rightarrow 0.00485\pm0.06746847118$

$\Rightarrow (0.00485-0.06746847118,\:0.00485 +0.06746847118)$

$\Rightarrow (-0.06261847118,\:0.07231847118)$

$\Rightarrow \mathbf{{\color{Blue} (-0.063,\:0.072)}}$

So the 99% confidence interval for $(p_{1}-p_{2})$ is calculated as $\color{Blue} (-0.063,\:0.072)$

The researchers are 99% confident that difference between the two population proportions, p₁ -p₂ , is between -0.063 and 0.072