Question

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 95% confidence interval for a difference in proportions p1-p2 if the samples have n1=70 with p^1=0.68 and n2=100 with p^2=0.56, and the standard error is SE=0.07. Round your answers to three decimal places. The 95% confidence interval is to .

Answer 1

Answer:

$(0.68-0.56) -1.96 \sqrt{\frac{0.68(1-0.68)}{70} +\frac{0.56*(1-0.56)}{100}}= -0.0263$

$(0.68-0.56) +1.96 \sqrt{\frac{0.68(1-0.68)}{70} +\frac{0.56*(1-0.56)}{100}}= 0.2663$

So then we are 95% confident that the true difference in the proportions is given by:

$-0.0263 \leq p_1 -p_2 \leq 0.2663$

Step-by-step explanation:

The information given for this case is:

$\hat p_1 = 0.68 , \hat p_2 = 0.56$

$n_1 = 70, n_2 =100$

We want to construct a confidence interval for the difference of proportions $p_1 -p_2$ and for this case this confidence interval is given by:

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

The confidence level is 0.95 so then the significance level would be $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ if we find the critical values for this confidence level we got:

$z_{\alpha/2}= \pm 1.96$

And replacing the info we got:

$(0.68-0.56) -1.96 \sqrt{\frac{0.68(1-0.68)}{70} +\frac{0.56*(1-0.56)}{100}}= -0.0263$

$(0.68-0.56) +1.96 \sqrt{\frac{0.68(1-0.68)}{70} +\frac{0.56*(1-0.56)}{100}}= 0.2663$

So then we are 95% confident that the true difference in the proportions is given by:

$-0.0263 \leq p_1 -p_2 \leq 0.2663$