Question

In a study of the length of time it takes to earn a master’s degree, 24 randomly-selected students had a mean of 2.4 years and a standard deviation of 0.8 years. Assume that the time to earn a master’s degree is normally distributed. You wish to construct a 90% confidence interval for the mean time it takes to earn a master’s degree. Use this information to answer the next 5 questions. (Be sure to use the correct distribution.)

What is the confidence interval? Round to 2 decimal places.

If applicable, what are the degrees of freedom? (Type NA if not applicable.)

What is the critical value?

What is the point estimate?

What is the margin of error? Round to 2 decimal places.

Answer 1

sample mean, xbar = 2.4
sample standard deviation, s = 0.8
sample size, n = 24
degrees of freedom, df = n - 1 = 23

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.714

ME = tc * s/sqrt(n)
ME = 1.714 * 0.8/sqrt(24)
ME = 0.28

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (2.4 - 1.714 * 0.8/sqrt(24) , 2.4 + 1.714 * 0.8/sqrt(24))
CI = (2.12 , 2.68)

b)

degrees of freedom, df = n - 1 = 23

c)

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.714

d)

point estimate = sample mean, xbar = 2.4

e)

ME = tc * s/sqrt(n)
ME = 1.714 * 0.8/sqrt(24)
ME = 0.28