Question

A random sample of 49 colleges yielded a mean cost of college education of $30,500 per year. Assume that the population standard deviation is $3000 and college costs are normally distributed.

a.) Calculate the standard error of the mean. Round to the hundredth.

b.) Calculate the lower bound for a 90% confidence interval for the population mean cost of education. Round to the hundredth.

c.) Calculate the upper bound for a 90% confidence interval for the population mean cost of education. Round to the hundredth

d.) You just constructed a 90% confidence interval for the population mean cost of education. How will the interval change if we increase the confidence to 99%?

e.) How large a sample is needed if we wish to estimate the population mean cost of college education within $200 with 90% confidence? Round answer up to the next whole number.

Answer 1

a)

standard error = 3000/sqrt(49) = 428.47

b)

sample mean, xbar = 30500
sample standard deviation, σ = 3000
sample size, n = 49

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.6449

ME = zc * σ/sqrt(n)
ME = 1.6449 * 3000/sqrt(49)
ME = 704.96

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (30500 - 1.6449 * 3000/sqrt(49) , 30500 + 1.6449 * 3000/sqrt(49))
CI = (29797.14 , 31202.86)

lower bound = 29797.14

c)

Upper bound = 31202.86

d)

sample mean, xbar = 30500
sample standard deviation, σ = 3000
sample size, n = 49

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 3000/sqrt(49)
ME = 1105.71

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (30500 - 2.58 * 3000/sqrt(49) , 30500 + 2.58 * 3000/sqrt(49))
CI = (29394.29 , 31605.71)

e)

The following information is provided,
Significance Level, α = 0.1, Margin or Error, E = 200, σ = 3000

The critical value for significance level, α = 0.1 is 1.64.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.64 * 3000/200)^2
n = 605.16

Therefore, the sample size needed to satisfy the condition n >= 605.16 and it must be an integer number, we conclude that the minimum required sample size is n = 606 or 605