Question

In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hired by the Better Business Bureau to investigate the complaints they received this year involving new car dealers. You plan to select a sample of new car dealer complaints to estimate the proportion of complaints the Better Business Bureau is able to settle. Assume the population proportion of complaints settled for new car dealers is .75, the same as the overall proportion of complaints settled in 2008
a. Suppose you select a sample of 450 complaints involving new car dealers. Show the sampling distribution of .
b. Based upon a sample of 450 complaints, what is the probability that the sample proportion will be within .04 of the population proportion?
c. Suppose you select a sample of 200 complaints involving new car dealers. Show the sampling distribution of .
d. Based upon the smaller sample of only 200 complaints, what is the probability that the sample proportion will be within .04 of the population proportion?
e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in part (b)?

Answer 1

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.