Chapter 10.1, Problem 23E

(a)

To determine

To explain does this study convincing evidence of a difference at the $\alpha =0.05$ significance level in the development of peanut allergies in infants like the ones in this study who consume and avoid peanut butter.

Answer to Problem 23E

There is convincing evidence that there is a difference in the development of peanut allergies in infants like the ones in this study who consumed or avoid peanut butter.

Explanation of Solution

Given:

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

Define Hypotheses:

So, the given claim that: difference in proportion.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

Where we have,

  $p_1=$ the proportion of infants who consumed a baby food form of peanut butter is allergic to peanuts.

  $p_2=$ the proportion of infants who avoided peanut butter is allergic to peanuts.

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,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{10}{307}\\ =0.03257\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{55}{321}\\ =0.17134\\ {\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{10+55}{307+321}\\ =\frac{65}{628}\\ =0.10350\end{array}$

Now, we will calculate the value of test statistics as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.03257-0.17134-0}{\sqrt{0.10350(1-0.10350)}\sqrt{\frac{1}{307}+\frac{1}{321}}}\\ \approx -5.71\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-5.71)\\ \approx 0\end{array}$

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence that there is a difference in the development of peanut allergies in infants like the ones in this study who consumed or avoid peanut butter.

(b)

To determine

To explain based on your conclusion in part (a), which mistake −Type I error or Type II error you could have made.

Answer to Problem 23E

Type I error.

Explanation of Solution

In part (a), the conclusion was made that there is convincing evidence that there is a difference in the development of peanut allergies in infants like the ones in this study who consumed or avoid peanut butter. And the hypothesis defined was that:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1\ne p_2\end{array}$

Thus, based on the conclusion in part (a), we have:

Type I error is the error when we reject the null hypothesis, when the null hypothesis is false.

And the Type II error is when we fail to reject the null hypothesis, when the null hypothesis is false.

Since in this case we rejected the null hypothesis, then it can mean that we have made a Type I error.

(c)

To determine

To explain should you generalize your results in part (a) to all infants or not and why.

Answer to Problem 23E

Thus, it cannot be generalize to all infants.

Explanation of Solution

As the results in part (a),

We can say that as the study only includes infants who had shown evidence of other kinds of allergies.

Then, it is possible that infants with some allergies are more likely to also have a peanut allergy. Thus, we cannot generalize the results to all infants as the proportion might be different for children who had not shown evidence of other kind of allergies.