Question

7.29 The mean preparation fee H&R Block charged retail customers last year 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 Excel to compute your answers.

Can you please help using excel I'm not understanding how to make this work with excel.

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. What is the minimum samples size if you want to have at least a .95 probability that the sample mean is within $8 of the population mean?

Answer 1

excel formulae are in bold:

a)

for normal distribution z score =(X-μ)/σx
mean μ=	183
standard deviation σ=	50

std error=σx̅=σ/√n=

9.1287

probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean =norm.dist(183+8,183,50/sqrt(30),true)-norm.dist(183-8,183,50/sqrt(30),true)

=0.6192

b)

probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean :

=norm.dist(183+8,183,50/sqrt(50),true)-norm.dist(183-8,183,50/sqrt(50),true)

=0.7421

c)

probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean

=norm.dist(183+8,183,50/sqrt(100),true)-norm.dist(183-8,183,50/sqrt(100),true)

=0.8904

d)

for95% CI crtiical Z          =	1.96	from excel:normsinv(0.975)
standard deviation σ=	50.000
margin of error E =	8
required n=(zσ/E)2 =(1.96*50/8)^2 =	151	Rounding up