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 3 percentage points with 96% confidence if he uses an estimate of 52% obtained from a poll?

(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?

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

(a) The results are close because the confidence 96% is close to 100%.

(b) The results are close because the margin of error 3% is less than 5%

(c) The results are close because 0.52(10.52)=0.2496 is very close to 0.25.

Explanation / Answer

Solution:-
a) p = 0.52 , E = 0.03 , Z = 2.054
n = (z/E)^2 * p * (1-p)
n = (2.054/0.03)^2*0.52*0.48
n = 1170.0460
n = 1170

b) p = 0.5

n = (2.054/0.03)^2*0.5*0.5
n = 1171.92
n = 1172

c) option (c) The results are close because 0.52(10.52)=0.2496 is very close to 0.25.

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