Question

Losing weight A Gallup poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the $95\%$confidence interval for the population

proportion who want to lose weight is $(0.56,0.62)$.

a. Interpret the confidence interval.

b. What is the point estimate that was used to create the interval? What is the margin of error?

c. Based on this poll, Gallup claims that more than half of U.S. adults want to lose weight. Use the confidence interval to evaluate this claim.

Short answer

a. There is $95\%$assurance that US adults who wish to lose weight lie in given interval.

b. Point estimation is $0.59$and margin of error is $0.03$.

c. It is a convincing evidence.

Step-by-step solution

Given Information

It is given that $95\%$confidence interval for proportion is$(0.56,0.62)$

Concept Used

Formula Used are$\hat{p}=\frac{\mathrm{UCL}+\mathrm{LCL}}{2}$and$\mathrm{ME}=\frac{\mathrm{UCL}-\mathrm{LCL}}{2}$

a. Interpreting Confidence Interval

It refers to $95\%$assurance that US adults proportion who wish to lose weight lie between$0.56-0.62$. Others do not wish to lose weight.

b. Point Estimate and Margin of Error

Point Estimation: It is given by $\hat{p}=\frac{0.56+0.62}{2}=0.59$

It lies halfway of confidence interval. Point estimate is average of boundaries of given interval.

Margin of Error: It is given by $\mathrm{ME}=\frac{0.62-0.56}{2}=0.03$

It is half the difference of confidence interval.

c. Evaluating the claim

After the above observation and given data, it can be observed that population proportion who wish to lose weight is $0.50$.

Hence, there is a convincing evidence that more than half of US adults who wish to lose weight.