Question

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo-controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to GI symptoms ("Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips," J. of the Amer. Med. Assoc., 1998: 150-152). Among 529 individuals in the TG control group, 17.6 % experienced an adverse GI event, whereas among the 563 individuals in the olestra treatment group, 15.8 % experienced such an event.

a. Carry out a test of hypotheses at the 5 % significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control treatment.

b. If the true percentages for the two treatments were 15 % and 20 %, respectively, what sample sizes

\((m = n\)) would be necessary to detect such a difference with probability 90?

Short answer

(a) There is not sufficient evidence to support the claim that the incidence rate of Gl problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group.

(b) \(n = 1211\)\(\)

Step-by-step solution

To determine the value of the test statistic

Given:

a)

\(\begin{array}{l}{{\hat p}_1} = 17.6\% = 0.176\\{n_1} = 529\\{{\hat p}_2} = 15.8\% = 0.158\\{n_2} = 563\\\alpha = 5\% = 0.05\end{array}\)

(a) Claim: \({p_1} \ne {p_2}\)

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states an equality. Since the null hypothesis is not the claim, the alternative hypothesis is the claim.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\{H_a}:{p_1} \ne {p_2}\end{array}\)

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}{x_1} = {{\hat p}_1}{n_1} = 0.176(529) \approx 93\\{x_2} = {{\hat p}_2}{n_2} = 0.158(563) \approx 89\end{array}\)

\({\hat p_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}} = \frac{{93 + 89}}{{529 + 563}} = \frac{1}{6} \approx 0.1667\)

The critical values are the values corresponding to a probability of \(0.025/0.975\) in table A.3:

\(z = \pm 1.96\)

The rejection region then contains all values below -1.96 and all values above 1.96.

Determine the value of the test statistic:

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }} = \frac{{0.176 - 0.158}}{{\sqrt {0.1667(1 - 0.1667)} \sqrt {\frac{1}{{529}} + \frac{1}{{563}}} }} \approx 0.80\)

If the value of the test statistic is within the rejection region, then the null hypothesis is rejected:

\( - 1.96 < 0.80 < 1.96 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that the incidence rate of Gl problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group.

To find the n value for the given sample sizes

(b) Given: Sample sizes are equal.

\(\begin{array}{l}{p_1} = 15\% = 0.15\\{p_2} = 20\% = 0.20\\P\left( {{\rm{ Reject }}{H_0}\mid {H_0}{\rm{ false }}} \right) = 0.90\end{array}\)

\({q_1}\)is the complement of \({p_1}\)and \({q_2}\)is the complement of \({p_2}\).

\(\begin{array}{l}{q_1} = 1 - {p_1} = 1 - 0.15 = 0.85\\{q_2} = 1 - {p_2} = 1 - 0.20 = 0.80\end{array}\)

\(\beta \)is the probability of not rejecting the null hypothesis when the null hypothesis \({H_0}\)is false (use the complement rule):

\(\beta = 1 - P\left( {{\rm{ Reject }}{H_0}\mid {H_0}{\rm{ false }}} \right) = 1 - 0.90 = 0.10\)

The critical value of the significance level \(\alpha \) is the value corresponding to a probability of \(\alpha /2 = 0.025\) in table A.3:

\({z_{\alpha /2}} = 1.96\)

The critical value of the probability of a type II error \(\beta \) is the value corresponding to a probability of 0.10 in table A.3:

\({z_\beta } = 1.28\)

\(d\)is the difference in population means:

\(d = {p_1} - {p_2} = 0.15 - 0.20 = - 0.05\)

Formula sample size:

Fill in the known values and evaluate (round up!):\(n = {\frac{{1.96\sqrt {(0.15 + 0.20)(0.85 + 0.80)/2} + 1.28\sqrt {0.15(0.85) + 0.20(0.80)} }}{{{{( - 0.05)}^2}}}^2} \approx 1211\)\(\)\(\)

Final proof

(a) There is not sufficient evidence to support the claim that the incidence rate of Gl problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group.

(b) \(n = 1211\)\(\)