Answer:
a) the sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.0204
b) the probability that the sample proportion will be within 0.04 of the population proportion is 0.95
c) sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.03061
d) the probability that the sample proportion will be within 0.04 of the population proportion is 0.8088
e) gain in precision is 0.1402.
Step-by-step explanation:
a) Let p represent the
Given that
population proportion of complaints settled for new car dealers p = 0.75.
and n = 450
mean of the sampling distribution of the sample proportion is the population proportion p
i.e up° = p
mean of the sampling distribution of the sample proportion p° = 0.75
so standard error of the proportion is;
αp° = √(p( 1-p ) / n)
we substitute
αp° = √(0.75 ( 1-0.75 ) / 450)
=√(0.1875 / 450
= √0.0004166
= 0.0204
therefore the sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.0204
b)
(p° - p) is within 0.04
so lets consider
p ( -0.04 ≤ p° - p ≤ 0.04) = p ( ( -0.04/√(0.75 ( 1-0.75 ) / 450)) ≤ z ≤ ( 0.04/√(0.75 ( 1-0.75 ) / 450))
= p( -0.04/0.0204 ≤ z ≤ 0.04/0.0204)
= p ( -1/96 ≤ z ≤ 1.96 )
= p( z < 1.96 ) - p( z < -1.96 )
now from the S-normal table,
area of the right of z = 1.96 = 0.9750
area of the left of z = - 1.96 = 0.0250
p( -0.04 ≤ p°- p ≤ 0.04) = p( z < 1.96 ) - p( z < -1.96 ) = 0.9750 - 0.0250
= 0.95
therefore the probability that the sample proportion will be within 0.04 of the population proportion is 0.95
c)
population proportion of complaints settled for new car dealers p = 0.75.
n = 200
mean of the sampling distribution of the sample proportion p°.
i.e up° = p
mean of the sampling distribution of the sample proportion p° = 0.75
Sampling distribution of the sample proportion p is determined as follows
αp° = √(p( 1-p ) / n)
we substitute
αp° = √(0.75 ( 1-0.75 ) / 200)
=√(0.1875 / 200
= √0.0009375
= 0.03061
therefore sampling distribution of the sample proportion is approximately normal with mean 0.75 and standard deviation is 0.03061
d)
(p° - p) is within 0.04
so lets consider
p ( -0.04 ≤ p° - p ≤ 0.04) = p ( ( -0.04/√(0.75 ( 1-0.75 ) / 200)) ≤ z ≤ ( 0.04/√(0.75 ( 1-0.75 ) / 200))
= p( -0.04/0.03061≤ z ≤ 0.04/0.03061)
= p ( -1.31 ≤ z ≤ 1.31 )
= p( z < 1.31 ) - p( z < -1.31 )
now from the S-normal table,
area of the right of z = 1.31 = 0.9049
area of the left of z = - 1.31 = 0.0951
p( -0.04 ≤ p°- p ≤ 0.04) = p( z < 1.31 ) - p( z < -1.31 ) = 0.9049 - 0.0951
= 0.8098
therefore the probability that the sample proportion will be within 0.04 of the population proportion is 0.8088
e)
From b), the sample proportion is within 0.04 of the population proportion; with the sample of 450 complaints involving new car dealers is 0.95.
sample proportion is within 0.04 of the population proportion; with the sample of 200 complaints involving new car dealers is 0.8098.
measured by the increase in probability, gain in precision occurs by taking the larger sample in part (b)
i.e
Gain in precision will be;
0.9500 − 0.8098
= 0.1402
therefore gain in precision is 0.1402.
