Question

Losing weight A Gallup Poll in November $2008$found that $59\%$of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is $\pm 3$percentage points.”

(a) Explain what the margin of error means in this setting.

(b) State and interpret the $95\%$ confidence interval.

(c) Interpret the confidence level.

Short answer

a). on average the sample proportion will be within $3\%$of the true proportion in $95\%$of all samples.

b). We are $95\%$confident that the true population proportion is between $0.56$and $0.62$.

c). On average, in $95\%$ of all samples the corresponding confidence interval will contain the true population proportion $p$.

Step-by-step solution

Part (a) Step 1: Given Information 

For the total sample of national adults, one can say with $95\%$confidence that the margin of (sampling) error is $\pm 3$ percentage points.

Part (a) Step 2: Explanation

The margin of error is the amount of error that is made, on average, in estimating the population proportion $p$ by the sample proportion $\hat{p}$.

Thus on average the sample proportion will be within $3\%$ of the true proportion in $95\%$ of all samples.

Part (b) Step 1: Given Information 

$\hat{p}=59\%=0.59$

Part (b) Step 2: Explanation 

The boundaries of the confidence interval is the sample proportion increased/decreased by the margin of error:

$$\begin{gathered} 0.56=0.59-0.03 \\ =\hat{p}-E<p<\hat{p}+E \\ =0.59+0.03 \\ =0.62 \end{gathered}$$

We are $95\%$confident that the true population proportion is between $0.56$and $0.62$.

Part (c) Step 1: Given Information

The confidence level is $95\%$.

Part (c) Step 2: Explanation 

The confidence level is $95\%$.

This means: On average, in $95\%$of all samples the corresponding confidence interval will contain the true population proportion $p$.