Question

A television sports commentator wants to estimate the proportion of citizens who "follow professional football." Complete parts (a) through (c). What sample size should be obtained if he wants to be within 4 percentage points with 96% confidence if he uses an estimate of 48% obtained from a poll? What sample size should be obtained if he wants to be within 4 percentage points with 96% confidence if he does not use any prior estimates? Why are the results from parts (a) and (b) so close?

Answer 1

Answer:

626 and 654

Step-by-step explanation:

Given that a  television sports commentator wants to estimate the proportion of citizens who "follow professional football."

Part I:

p = 0.48

$q=1-p = 0.52\\se = \sqrt{\frac{pq}{n} } \\\frac{0.4996}{\sqrt{n} } =$

Margin of error =$2.045*\frac{0.4996}{\sqrt{n} }<0.04\\n>626$

Sample size should be >626

Part II:

If unknown we take p = 0.5 because maximum std error for this

Here everything would be the same except insted of 0.48 we use 0.5

Margin of error = $2.045*\sqrt{\frac{0.5}{\sqrt{n} } } <0.04\\n>654$

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a and b are too close because 0.46 proportion is close to part b proporti0n 0.5

Answer 2

The sample sizes obtained for the proportion to be within 4 percentage points with 96% confidence are respectively; 626 and 654

What is the sample size?

A) We are given;

Sample proportion; p = 48% = 0.48

Thus; q = 1 - p = 0.52

Margin of error = 4% = 0.04

Formula for margin of error is;

M = z√(p(1 - p)/n)

z at 96% CI is 2.045. Thus

2.045√(0.48 * 0.52/n) = 0.04

Solving for n gives n = 626

B) If he does not want to use any prior estimate, then it means that;

Sample proportion; p = 0.5

Thus;

2.045√(0.5 * 0.5/n) = 0.04

solving for n gives; n = 654

C) The results from parts A and B are close because their sample proportions are close.

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