Q.12 - The Practice of Statistics for AP

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. 

Step-by-Step Solution

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 $p$.

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 $p$.