Question

Preventing peanut allergies Refer to Exercise 23.

a. Construct and interpret a $95\%$ confidence interval for the difference between the true proportions. Assume that the conditions for inference are met.

b. Explain how the confidence interval provides more information than the test in

Exercise 23.

Short answer

a. Confidence Interval is $(0.09302,0.18452)$

b. The significance test checks the one possible value for the difference between the proportions.

Step-by-step solution

Given Information

It is given that $x_1=10$

$x_2=55$

$n_1=307$

$n_2=321$

$\alpha =0.05$

Confidence Interval

All the three conditions random, independent and Normal are satisfied.

Hence, sample proportion ${\hat{p}}_1=\frac{x_1}{n_1}=\frac{10}{307}=0.03257$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{55}{321}=0.17134$

For $1-\alpha =0.95$using table, $z_{a/2}=1.96$

Confidence Interval is $\left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.03257-0.17134)-1.96\times \sqrt{\frac{0.03257(1-0.03257)}{307}+\frac{0.17134(1-0.17134)}{321}}$

$\approx -0.18452$

and $\left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}\times \sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}$

$=(0.03257-0.17134)+1.96\times \sqrt{\frac{0.03257(1-0.03257)}{307}+\frac{0.17134(1-0.17134)}{321}}$

$\approx -0.09302$

Confidence Interval is$\left(-0.18452,-0.09302\right)$

How confidence interval provides more information than the test in previous exercise.

As compared to significance test, confidence interval gives more information as it gives range of values and significance test checks one possible value of difference between proportions.