Question

The population proportion is .80. What is the probability that a sample proportion will be within...

The population proportion is .80. What is the probability that a sample proportion will be within +/- .03 of the population proportion for each of the following sample sizes? Round your answers to 4 decimal places. Use z-table.

a. n=100
b. n=200
c. n=500
d. n=1000
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Answer #1

It is known that,

for large sample sizes (n > 30),

p~ Normal(0.8,0(1-0.8)

a) If n = 100,

p ~ Normal(0.8,0.8*(1-0.8) 100 or ~Normal (0.8,0.042)

Now,

(lp_ 0.81 0.03)-P(P - 0.81 〈 0.03,

P 0.80.03) - P(Z 0.75)

Plp-0.8] < 0.03)-P(Z < 0.75) _ P(Z <-0.75)

Plpー0.8] < 0.03)-Φ(0.75) _ Φ(-0.75)

Plpー0.8! < 0.03) 0.773373-0.226627

Plp-0.8] < 0.03) 0.546746

b) If n = 200,

p ~ Normal(o.8, 0.8*(1-0.8) 0.8(1 0.8) 200 or ~ Normal (0.8,0.02828432)

Now,

P( 0.8 003)-0.80.03 Plp-0.8| < 0.03) P( 4 0.02828430.0282843 ) = P(|Z| < 1.06066) - , ,

Plp_ 0.8] < 0.03)-P(Z < 1.06066)-P(Z <-1.06066)

Plpー0.8] < 0.03)-Φ(1.06066) _ Φ(-1.06066)

Plp-0.8] < 0.03) 0.855578-0.144422

Plp-0.8] < 0.03) 0.711156

c) If n = 500,

hat{p} ~ Normal(0.8,rac{0.8*(1-0.8)}{500}) or Normal(0.8, 0.01788852)

Now,

0.03 Pllp_ 0.81S 0.03) = P(上ー0.81 < 4 0.01788850.0178885 ) = P(21 £1.67705)

Plp_ 0.8] < 0.03)-P(Z < 1.67705)-P(Z <-1.67705)

Plp-0.8] < 0.03) 0.953234-0.0467664

Plp-0.8] < 0.03) 0.906468

d) If n = 1000,

p ~ Normal(0.8, 0.8*(1-0.8) 1000 or p~ Normal (0.8,0.01264912)

Now,

Plp-0.8| < 0.03) 1-0.81·0.03 P( 4 ) = P(|Z| < 2.37171) - , 0.0126491-0.0126491

Plp-0.8] < 0.03)-P(Z < 2.37171)-P(Z <-237171) 271 71

Plp-0.8] < 0.03) 0.99 1147-0.0088530

Plp-0.8] < 0.03) 0.982294

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