Question

A television sports commentator wants to estimate the proportion of citizens who​ "follow professional​ football." complete parts​ (a) through​ (c).

Answer 1

Part A

The number of samples needed to get a confidence interval with a margin of error M is given by:

$n= \frac{z_{\alpha/2}^2p(1-p)}{M^2}$

where $z_{\alpha/2}$ is the z-score of the confidence level and p is the population proportion.

If he wants to be within 4 percentage points with 96% confidence and he uses an estimate of 48% obtained from a poll, the sample size that should be obtained is given by:

$n= \frac{(2.054)^2\times0.48(1-0.48)}{(0.04)^2} \\ \\ = \frac{4.218916\times0.48(0.52)}{0.0016} = \frac{1.053041434}{0.0016} \\ \\ =\lceil658.15\rceil=659$



Part B:

If he wants to be within 4 percentage points with 96% confidence and he does not use any prior estimates, the sample size that should be obtained is given by:

$n= \frac{(2.054)^2\times0.50(1-0.50)}{(0.04)^2} \\ \\ = \frac{4.218916\times0.50(0.50)}{0.0016} = \frac{1.054729}{0.0016} \\ \\ =\lceil659.21\rceil=660$



Part C:

The resulta from parts a and b are close because the result from the poll (i.e. 48%) is close to the conservative proportion used when there is no prior knowledge of any proportion (i.e. 50%).