Question

Prayer in school A New York Times/CBS News Poll asked the question, “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Sixty-six percent of the sample answered “Yes.” The article describing the poll says that it “is based on telephone interviews conducted from Sept. 13 to Sept. 18 with $1664$adults around the United States, excluding Alaska and Hawaii. . . . The telephone numbers were formed by random digits, thus permitting access to both listed and unlisted residential numbers.” The article gives the margin of error for a $95\%$confidence level as $3$percentage points.

(a) Explain what the margin of error means to someone who knows little statistics.

(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 localid="1649754733996" $95\%$confident that the true population proportion is between localid="1649754744695" $0.63$and localid="1649754756510" $0.69$.

c). On average, in localid="1649754767371" $95\%$of all samples the corresponding confidence interval will contain the true population proportion localid="1649754793252" $p$.

Step-by-step solution

Part (a) Step 1: Given Information 

The article gives the margin of error for a $95\%$confidence level as $3$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 

The article gives the margin of error for a $95\%$.

$\hat{p}=66\%=0.66$

$E=3\%=0.03$

Part (b) Step 2: Explanation 

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

$0.63=0.66-0.03=\hat{p}-E<p<\hat{p}+E=0.66+0.03=0.69$

We are $95\%$confident that the true population proportion is between $0.63$and$0.69$.

Part (c) Step 1: Given Information 

The article gives the margin of error for a $95\%$confidence level.

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 $p$.