Q. 41 - The Practice of Statistics for AP

Question

A TV poll A television news program conducts a call-in poll about a proposed city ban on handgun ownership. Of the $2372$ calls, $1921$ oppose the ban. The station, following recommended practice, makes a confidence statement: $"81\%$of the Channel $13$ Pulse Poll sample opposed the ban. We can be $95\%$ confident that the true proportion of citizens opposing a handgun ban is within$1.6\%$ of the sample result."

(a) Is the station's quoted $1.6\%$ margin of error correct? Explain.

(b) Is the station's conclusion justifed? Explain.

Step-by-Step Solution

Answer

a) No

b) No

Part (a) Step1 : Given Information

Need to explain the margin of error quoted is $1.6\%$

Part (a) Step2: Explanation

No, because the sample is a voluntary response sample, the confidence interval method's randomness condition is not met, and so the claimed $1.6$ percent margin of error is wrong.

Part (b) Step 2: Given Information

Need to explain the stations conclusion.

Step 5: Explanation(Part b)

No, because the sample is a voluntary response sample, the confidence interval method's randomization condition is not met, and hence the conclusion based on the confidence interval is erroneous.