Chapter 10.1, Problem 25E

(a)

To determine

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

Answer to Problem 25E

There is $95\%$ confidence that the true proportion of infants who consumed a baby food form of peanut butter and is allergic to peanuts is between $0.09302\text{ and 0}\text{.18452}$ higher than the true proportion of infants who avoided peanut butter and is allergic to peanuts.

Explanation of Solution

Referring to the previous exercise we have,

Given:

  $\begin{array}{l}{x}_1=10\\ x_2=55\\ n_1=307\\ n_2=321\\ \alpha =0.05\end{array}$

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the infants are independent random samples.

Independent: It is satisfied because the $307$ infants are less than $10\%$ of all infants and $321$ infants are less than $10\%$ of all infants.

Normal: It is satisfied because there are $10$ successes and $307-10=297$ failures in the first sample and there are $55$ successes and $321-55=266$ failures in the second sample, which are all at least ten.

Thus, all the conditions are satisfied.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{10}{307}\\ =0.03257\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{55}{321}\\ =0.17134\end{array}$

Now, for the confidence interval $1-\alpha =0.95$ , we have to find out the value of $z_{\alpha /2}=z_{0.025}$ using the table. Thus the z score will be:

  $z_{\alpha /2}=1.96$

Thus, the confidence interval will be:

  $\begin{array}{l}({\widehat{p}}_1-{\widehat{p}}_2)-z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.03257-0.17134)-1.96\times \sqrt{\frac{0.03257(1-0.03257)}{307}+\frac{0.17134(1-0.17134)}{321}}\\ \approx -0.18452\\ ({\widehat{p}}_1-{\widehat{p}}_2)+z_{\alpha /2}\times \sqrt{\frac{{\widehat{p}}_1(1-{\widehat{p}}_1)}{n_1}+\frac{{\widehat{p}}_2(1-{\widehat{p}}_2)}{n_2}}\\ =(0.03257-0.17134)+1.96\times \sqrt{\frac{0.03257(1-0.03257)}{307}+\frac{0.17134(1-0.17134)}{321}}\\ \approx -0.09302\end{array}$

Thus, we conclude that, we are $95\%$ confident that the true proportion of infants who consumed a baby food form of peanut butter and is allergic to peanuts is between $0.09302\text{ and 0}\text{.18452}$ higher than the true proportion of infants who avoided peanut butter and is allergic to peanuts.

(b)

To determine

To explain how the confidence interval provides more information than the test in previous exercise.

Explanation of Solution

The confidence interval gives more information than a significance test because the confidence interval gives us a range of possible values for the difference between the true proportions and the significance test only checks the one possible value for the difference between the proportions.