Question

Genetic Binge Eating. According to an article in Science News, binge eating has been associated with a mutation of the gene for a brain protein called melanocortin 4 receptor (MC4R). In one study, F. Horber of the Hirslanden Clinic in Zurich and his colleagues genetically analyzed the blood of 469 obese people and found that 24 carried a mutated MCAR gene. Suppose that you want to estimate the proportion of all obese people who carry a mutated MC4R gene

a. Determine the margin of error for a $90\%$confidence interval.

b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a $95\%$confidence interval. Explain your answer

Short answer

(a) the margin of error for a $90\%$confidence interval is $0.0166$

(b) A$90\%$confidence interval has a lower margin of error than a $95\%$confidence interval.

Step-by-step solution

Part (a) Step 1: Given Information 

Given in the question that, according to an article in science news binge has been associated with a mutation of the gene for a brain protein called melanocortin 4 receptor. In one study, F.Horber of the Hirslanden clinic in Zurich and his colleagues genetically analyzed the blood of 469 obese people and found that 24 carried a mutated MC4R gene. Suppose that you want to estimate the proportion of all obese people who carry a mutated MC4R gene. we have to Determine the margin of error for a $90\%$confidence interval .

Part (a) Step 2: Explanation 

The formula of margin of error :$E=z_a\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$$\begin{gathered} \hat{p}=\frac{x}{n} \\ =\frac{24}{469} \\ \approx 0.05 \end{gathered}$$

Level of significance $\alpha =0.10$,$zvalueis1.65$

The margin of error for $90\%$confidence interval is calculated as folloes:

$$\begin{gathered} E=z_a\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =1.65\sqrt{\frac{0.05(1-0.05)}{469}} \\ =0.0166 \end{gathered}$$

Part (b) Step 1: Given Information 

We have to calculate Without doing any calculations, indicate whether the margin of error is larger or smaller for a $95\%$confidence interval.

Part (b) Step 2: Explanation 

When the level of confidence rises, the related critical value rises as well. The margin of error is the product of the statistic's critical value and standard error. As a result, as the level of confidence rises, so does the margin of error. Furthermore, the confidence interval grows bigger. In addition, as the level of confidence drops, the margin of error diminishes.

As a result, $90\%$confidence interval has a lower margin of error than a $95\%$confidence interval