Question

Prayer in school A New York Times/CBS News Poll asked a random sample of U.S. adults the question “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the $95\%$confidence interval for the population proportion who favor such an amendment is $(0.63,0.69)$.

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, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.

Short answer

a. $95\%$confidence interval for the population is given by $(0.63,0.69)$

b. The margin of error is $0.03$

c. There is no evidence that $0.67$US Adults favour amendment.

Step-by-step solution

Given Information

Confidence Interval is $95\%$

Population proportion is$\left(0.63,0.69\right)$

Concept Used

Formulas $\hat{p}=\frac{\mathrm{UCL}+\mathrm{LCL}}{2}$

$\mathrm{ME}=\frac{\mathrm{UCL}-\mathrm{LCL}}{2}$

a. Interpreting Confidence Level

It refers to $95\%$assurance that population proportion who favor such an amendment is $\left(0.63,0.69\right)$.

Others do not favour amendment do not come in this range.

b. Point Estimate and Margin of Error

Proportion is $\hat{p}=\frac{0.63+0.69}{2}$

and Margin of Error $ME=\frac{0.69-0.63}{2}=0.03$

Point Estimate: Since it lies halfway of confidence interval, point estimate is average of boundaries of interval. Here point estimate of $\left(0.63,0.69\right)$is $0.66$

Margin of Error: It is half the difference of confidence interval.

c. Evaluating the Claim

From the data and above calculations, it is observed that confidence interval contains $\frac{2}{3}=0.66$.

It means $0.67$of US adults favours amendment.

Hence, it is not a convincing evidence.