Question

The mean preparation fee H&R Block charged retail customers in 2012 was $183 (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is $50. Use z-table here (link below). Round to four decimal places.

a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean?
b. What is the probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean?
c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean?
d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a 0.95 probability that the sample mean is within $8 of the population mean?

Answer 1

a)0.6192

b)0.7422

c)0.8904

d)at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.

Explanation:

Let z(p) be the z-statistic of the probability that the mean price for a sample is within the margin of error. Then

z(p)=
`(ME*√(N))/(s )` where

• Me is the margin of error from the mean

• s is the standard deviation of the population

• N is the sample size

a.

z(p)=
`(8*√(30))/(50 )` ≈ 0.8764

by looking z-table corresponding p value is 1-0.3808=0.6192

b.

z(p)=
`(8*√(50))/(50 )` ≈ 1.1314

by looking z-table corresponding p value is 1-0.2578=0.7422

c.

z(p)=
`(8*√(100))/(50 )` ≈ 1.6

by looking z-table corresponding p value is 1-0.1096=0.8904

d.

Minimum required sample size for 0.95 probability is

N≥
`((z*s)/(ME) )^2` where

• N is the sample size

• z is the corresponding z-score in 95% probability (1.96)

• s is the standard deviation (50)

• ME is the margin of error (8)

then N≥
`((1.96*50)/(8) )^2` ≈150.6

Thus at least 151 sample is needed for 95% probability that sample mean falls within 8$ of the population mean.