Question

A random sample of 50 students has a mean GPA of 2.55. It is known that the population standard deviation for the GPA of all students is 1.1. Use this information to complete the following:

a) Does the information above provide an estimate of a population mean or a population proportion?

b) Provide a point estimate for the population mean GPA of all students.

c) Construct and interpret a 95% confidence interval for the parameter you selected in part (a).

d) What is the margin of error of this interval estimate?

e) How large of a sample would you need if you wanted your 95% confidence interval to be within 0.2 units of the population parameter?

Answer 1

Given, X bar = 2.55, S = 1.1 and n = 50

a)

the information above provide an estimate of a population mean

b)

point estimate for the population mean GPA of all students = 2.55

c)

95% confidence interval

α = 0.05 and df = 49

tc = 2.0096

CI = X bar +/- tc * S/SQRT(n)

CI = 2.55 +/- 2.0096*1.1/SQRT(50)

= 2.55 +/- 0.3126

= (2.2374, 2.8626)

d)

margin of error of this interval estimate = tc * S/SQRT(n) = 2.0096*1.1/SQRT(50) = 0.3126

e)

ME = 0.2

α=0.05, Zc​ = 1.96.

n >= (Zc*S/ME)^2

= (1.96*1.1/0.2)^2

= 116.2

>= 117 (Nearest value)