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?
Given, X bar = 2.55, S = 1.1 and n = 50
the information above provide an estimate of a population mean
point estimate for the population mean GPA of all students = 2.55
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)
margin of error of this interval estimate = tc * S/SQRT(n) = 2.0096*1.1/SQRT(50) = 0.3126
ME = 0.2
α=0.05, Zc = 1.96.
n >= (Zc*S/ME)^2
>= 117 (Nearest value)
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