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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 \(\mathrm{TG}\) 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 ?

Step by step solution

01

Define Hypotheses

We need to establish our null and alternative hypotheses. For this problem, the null hypothesis \(H_0\) is that the proportion of GI symptoms in the olestra group (\(p_1\)) is equal to the proportion in the TG control group (\(p_2\)). Thus, \(H_0: p_1 = p_2\). The alternative hypothesis \(H_a\) is that the proportions are different: \(H_a: p_1 eq p_2\).

02

State Significance Level

The significance level \(\alpha\) is given as 5%, or 0.05. This means we will reject the null hypothesis if the p-value is less than 0.05.

03

Calculate Test Statistic

Calculate the test statistic using the two-proportion z-test formula: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] where \( \hat{p}_1 = 0.158, \hat{p}_2 = 0.176\), \( n_1 = 563, n_2 = 529 \), and \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \) where \( x_1 = 0.158 \times 563 \) and \( x_2 = 0.176 \times 529 \).

04

Pooled Proportion Calculation

First, calculate \( \hat{p} \). This is the combined proportion of GI incidents: \( \hat{p} = \frac{(0.158 \times 563) + (0.176 \times 529)}{563 + 529} \).

05

Test Statistic Value

Plug the values of \( \hat{p}, \hat{p}_1, \hat{p}_2, n_1, \text{and } n_2 \) into the z-test formula and compute the z-value.

06

Determine Critical Value & Decision Rule

For a two-tailed test at \( \alpha = 0.05 \), the critical values correspond to z-scores of approximately \( \pm 1.96 \). If the computed z-value is beyond these critical values, we reject \( H_0 \).

07

Conclusion

Compare the computed z-value with the critical values. If it falls beyond \( \pm 1.96 \), reject the null hypothesis, indicating a statistically significant difference. Otherwise, do not reject \( H_0 \).

08

Sample Size Calculation (Power Analysis)

Now, for part (b), use the formula for sample size calculation with power \( 0.9 \). The formula is: \[ n = \left( \frac{Z_{1-\alpha/2} + Z_\beta}{p_1 - p_2} \right)^2 \cdot \left( p_1(1-p_1) + p_2(1-p_2) \right) \] where \(Z_{1-\alpha/2}\) is the z-score for \(\alpha/2 = 0.025\) and \(Z_\beta\) is the z-score for \(1 - \beta = 0.90\). Use \( p_1 = 0.15 \), \( p_2 = 0.20 \).

09

Calculate Required Sample Size

Substitute the values into the sample size formula to compute \(n\), the required sample size for each group to achieve 90% power.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-Proportion Z-Test

When comparing the proportions of two groups, like those experiencing gastrointestinal (GI) symptoms from olestra and regular chips, a two-proportion z-test is a useful statistical method. This test checks if there's a significant difference between the two group proportions.

In our example, we define the null hypothesis (H_0) as there being no difference (p_1 = p_2) between the proportions of GI symptoms in the two groups. The alternative hypothesis (H_a) suggests a difference does exist (p_1 eq p_2). The z-test statistic formula helps us understand if the observed difference in sample proportions is large enough to be significant:

\[z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]

Here, \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes, and \(\hat{p}\) is the pooled proportion. This pooled proportion is calculated by combining the incidences of GI symptoms in both groups.

Significance Level

In hypothesis testing, the significance level (\(\alpha\)) helps determine whether to reject the null hypothesis. It represents the probability of making a type I error—rejecting \(H_0\) when it is actually true.

For our problem, the significance level is set at 5% or 0.05. This value is a threshold for deciding if the differences we observe between the olestra and regular chip groups are statistically significant. If the p-value, calculated from the z-test, is less than 0.05, we reject \(H_0\). But if it's more, we don't have enough evidence to say the groups are different.

In simpler terms, a 0.05 significance level suggests that there's a 5% chance that any difference we've observed is purely due to random chance, rather than a true difference in the population.

Sample Size Calculation

Calculating the right sample size is crucial in designing an experiment. It ensures we have enough data to detect a significant effect if it exists. For the chewy dilemma of olestra vs. regular chips, the sample size calculation involves using the anticipated effect and the set significance level and power.

To find out how many people need to be in each group, we use the formula:

\[ n = \left(\frac{Z_{1-\alpha/2} + Z_{\beta}}{p_1 - p_2}\right)^2 \cdot \left( p_1(1-p_1) + p_2(1-p_2) \right) \]

Here, \(p_1\) and \(p_2\) are the assumed true proportions for the two treatments. \(Z_{1-\alpha/2}\) is the critical z-value for the significance level, and \(Z_{\beta}\) is for the power level. This formula lets you find the necessary sample size for each group to detect a difference with the desired accuracy.

Power Analysis

In any experiment, achieving the right power is vital. Power analysis helps to determine the probability of correctly rejecting the null hypothesis when it is false (avoiding a Type II error). It is essentially about ensuring that if there is a real effect, our test will detect it most of the time.

For this experiment, we desire a 90% power level. This means there's a 90% chance we'll detect a difference between the two chip types if it truly exists. Power is influenced by several factors, including the effect size (difference between proportions \(p_1\) and \(p_2\)), significance level, and sample size.

The larger the sample size, the greater the power. By setting a high power at the start, researchers ensure that their study is robust enough to detect meaningful differences and provide reliable results.