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 0.90?

Answer 1

Solution

It is given that 17.6% of the individuals from the TG group experienced a GI event from a group of 529 individuals and 15.8% of the individuals from the Olestra treatment group experienced a GI event from a group of 563 individuals.

Let ${\hat{p}}_1$ denote the sample proportion of individuals from the TG group who experienced the GI event and ${\hat{p}}_2$ denote the sample proportion of individuals from the Olestra treatment group who experienced the GI event.

Thus,

$\begin{array}{cc}{\hat{p}}_1& =0.176\\ {\hat{p}}_2& =0.158\end{array}$

The sample sizes for the TG group individuals and the Olestra treatment group are $n_1=529,n_2=563$, respectively.

Thus, the combined proportion is computed as follows:

$\begin{array}{cc}& \hat{p}=\frac{n_1{\hat{p}}_1+n_2{\hat{p}}_2}{n_1+n_2}\\ & =\frac{529\times 0.176+563\times 0.158}{529+563}\\ & =\frac{93.104+88.954}{1092}\\ & =\frac{182.058}{1092}\\ & =0.167\end{array}$

Part a:

The test hypotheses are as follows:

$\begin{array}{cc}{H}_0& :p_1=p_2\\ H_1& :p_1\ne p_2\end{array}$

Here, $p_1$ and $p_2$ denote the population proportion for the TG group and the Olestra treatment group,respectively.

The test statistic is computed as follows:

$\begin{array}{cc}Z& =\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}\\ & =\frac{0.176-0.158}{\sqrt{0.167(1-0.167)\left(\frac{1}{529}+\frac{1}{563}\right)}}\\ & =\frac{0.018}{0.022584}\\ & =0.7970\end{array}$

The test statistic value is $Z=0.7970$.

The $Z$-critical value with the level of significance $\alpha =0.05$ is obtained as 1.96 from the standard normal table.

The null hypothesis is rejected if $Z$-value > $Z$-critical value; otherwise, the null hypothesis fails to be rejected.

Since $Z$-value< $Z$-critical value, the null hypothesis fails to be rejected.

Hence, it can be concluded that the incidence rate of GI problems for the two groups is the same.

Part b:

The true proportions for the two treatments are given as follows:

$\begin{array}{cc}{p}_1& =0.15\\ p_2& =0.20\end{array}$

The Z-critical value with level of significance 0.10 is 1.64. Let the power be 80%, thus $\beta =0.20$. Hence, the $Z$-value with 0.20 is 0.84.

Thus, the sample size is computed as follows:

$\begin{array}{cc}& n={\left(Z_{\frac{\alpha }{2}}+Z_{\beta }\right)}^2\times \frac{p_1(1-p_1)+p_2(1-p_2)}{{\left(p_1-p_2\right)}^2}\\ & ={(1.96+0.84)}^2\times \frac{0.15(1-0.15)+0.20(1-0.20)}{{\left(0.15-0.20\right)}^2}\\ & ={2.48}^2\times \frac{0.1275+0.16}{-{0.05}^2}\\ & =6.1504\times 115\\ & =707.296\\ & \approx 708\end{array}$