A television sports commentator wants to estimate the proportion of citizens who "follow professional football." Complete parts (a) through (c).
Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
(a) What sample size should be obtained if he wants to be within 3 percentage points with $96 \%$ confidence if he uses an estimate of $54 \%$ obtained from a poll?
The sample size is $1165^{\prime}$. (Round up to the nearest integer.)
(b) What sample size should be obtained if he wants to be within 3 percentage points with $96 \%$ confidence if he does not use any prior estimates?
The sample size is 1168 (Round up to the nearest integer)
(c) Why are the results from parts (a) and (b) so close?
A. The results are close because $0.54(1-0.54)=0.2484$ is very close to 0.25
B. The results are close because the margin of error $3 \%$ is less than $5 \%$.
C. The results are close because the confidence $96 \%$ is close to $100 \%$.
Answer
For part (c), the results from parts (a) and (b) are close because the estimated proportions used in parts (a) and (b) are close to each other. Specifically, \(0.54 \cdot (1-0.54) = 0.2484\) is very close to \(0.5 \cdot (1-0.5) = 0.25\). Therefore, the sample sizes calculated using these proportions will also be close. The correct answer is \(\boxed{0.54(1-0.54)=0.2484}\) is very close to 0.25.
Step 1 ：A television sports commentator wants to estimate the proportion of citizens who follow professional football. He wants to be within 3 percentage points with 96% confidence.
Step 2 ：For part (a), he uses an estimate of 54% obtained from a poll. The sample size needed is calculated using the formula \(n = \lceil \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \rceil\), where \(Z = 2.05\) is the z-score for 96% confidence, \(E = 0.03\) is the margin of error, and \(p = 0.54\) is the estimated proportion. This gives a sample size of \(\boxed{1160}\).
Step 3 ：For part (b), he does not use any prior estimates. The worst-case scenario for the estimated proportion is used, which is \(p = 0.5\). The same formula is used to calculate the sample size, giving a result of \(\boxed{1168}\).
Step 4 ：For part (c), the results from parts (a) and (b) are close because the estimated proportions used in parts (a) and (b) are close to each other. Specifically, \(0.54 \cdot (1-0.54) = 0.2484\) is very close to \(0.5 \cdot (1-0.5) = 0.25\). Therefore, the sample sizes calculated using these proportions will also be close. The correct answer is \(\boxed{0.54(1-0.54)=0.2484}\) is very close to 0.25.
