Question

A television sports commentator wants to estimate the proportion of citizens who follow professional football. Complete part
Standard Normal Distribution Table (page 2) Standard Normal Distribution Table (page 1) SN Distri Standard Nurul Distre LA 0.
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Answer #1

Solution:

a)

Given,

E = 4% =  0.04

c = 96% = 0.96

\hat p or p = 52% = 0.52

1 - \hat p 1- p = 1 - 0.52 = 0.48

Now,

\alpha = 1 - c = 1 - 0.96 = 0.04

\alpha/2 = 0.02

\thereforea/2 = 2.054 (using z table)

The sample size for estimating the proportion is given by

n = 2/2P(1 - p) E2

= (2.054)2 * 0.52 * 0.48/ (0.042)

= 658.150896

= 659 ..(round to the next whole number)

n = 659

b)

Given,

E = 4% =  0.04

c = 96% = 0.96

here no any prior estimate is given si we it as p = 0.5

\hat p or p = 0.5

1 - \hat p = 1- p = 1 - 0.5 = 0.5

Now,

\alpha = 1 - c = 1 - 0.96 = 0.04

\alpha/2 = 0.02

\thereforea/2 = 2.054 (using z table)

The sample size for estimating the proportion is given by

n = 2/2P(1 - p) E2

= (2.054)2 * 0.5 * 0.5/ (0.042)

= 659.205625

= 660 ..(round to the next whole number)

n = 660

c)

The results are so close because,

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

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