Question

In each of the following cases, find the mean, variance, and standard deviation of the sampling distribution of the sample proportion .

(a) p = .5, n = 259 (Round variance to 6 decimal places and standard deviation to 4 decimal places. Do not round intermediate values.)

(b)

p = .2, n = 124 (Round variance to 6 decimal places and standard deviation to 4 decimal places. Do not round intermediate values.)

(c)

p = .8, n = 388 (Round variance to 6 decimal places and standard deviation to 4 decimal places. Do not round intermediate values.)

(d)	p = .96, n = 1,032 (Round variance to 6 decimal places and standard deviation to 4 decimal places. Do not round intermediate values.)

(c)

p = .8, n = 388 (Round variance to 6 decimal places and standard deviation to 4 decimal places. Do not round intermediate values.)

Answer 1

a)

Mean = p = 0.5

Variance = p( 1 - p) / n  = 0.5 * 0.5 / 259 = 0.000965

Standard deviation = sqrt ( Variance) = sqrt ( 0.000965) = 0.0311

b)

Mean = p = 0.2

Variance = p ( 1 - p) / n = 0.2 * ( 1 - 0.2) / 124 = 0.001290

Standard deviation = sqrt ( Variance) = sqrt (0.001290 ) = 0.0359

c)

Mean = p = 0.8

Variance = p ( 1 - p) / n = 0.8 * ( 1 - 0.8) / 388 = 0.000412

Standard deviation = sqrt ( Variance) = sqrt (0.000412) = 0.0203

d)

Mean = p = 0.96

Variance = p ( 1 - p) = 0.96 * ( 1 - 0.96) / 1032 = 0.000037

Standard deviation = sqrt ( Variance) = sqrt (0.000037) = 0.0061