A television sports commentator wants to estimate the proportion of citizens who

Question

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

?(a) What sample size should be obtained if he wants to be within 33 percentage points with 94?% confidence if he uses an estimate of 52?% obtained from a? poll? The sample size is nothing. ?(Round up to the nearest? integer.)?

(b) What sample size should be obtained if he wants to be within 33 percentage points with 94?% confidence if he does not use any prior? estimates?The sample size isnothing. (Round up to the nearest? integer.)

?(c) Why are the results from parts? (a) and? (b) so? close?

A. The results are close because the margin of error 33?% is less than? 5%.

B. The results are close because 0.52(1?0.52)=0.2496 is very close to 0.25.

C. The results are close because the confidence 94?% is close to? 100%.

Explanation / Answer

a)

n = z^2 pq/e^2

for 94 % z = normsinv(0.97) = 1.880793608

n = 1.880793608^2 * 0.52 * 0.48/0.03^2
= 981.03466

hence n = 982

b)
when estimate for p is not known
n = z^2/(4e^2)
=1.880793608^2 /(4*0.03^2)
982.6068
hence
n = 983
c)
B. The results are close because 0.52(1?0.52)=0.2496 is very close to 0.25.
is correct

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