Question

⦁   Given the following GPA for 4 students: 2, 3, 3, 4. Age of students are: 10, 11, 12, 15. There are 10,000 total students. Average GPA: Unknown, Standard Deviation of GPA: 0.02 and it’s Normal.

⦁   Find the mean and standard deviation for GPA. (10 points)

⦁   Calculate a 95% confidence interval for the populations mean GPA. (5 points)

⦁   Use Correlation formula to find r between GPA and Age. (10 points)

⦁   Find Regression Line Equation. (5 points)

⦁   If a student’s age is 13, what is the corresponding GPA? (5 points)

Answer 1

Mean GPA = (2 + 3 + 3 + 4)/4 = 3

The standard deviation of GPA = ((2 - 3)^2 + (3 - 3)^2 + (3 - 3)^2 + (4 - 3)^2)/4 - 1 = 0.82

The 95% confidence interval for the populations mean GPA is:

= 3 +- 3.18*(0.82/4)

= 1.70, 4.30

r = 0.945

The regression line is:

y = -1.29 + 0.36*x

Put x = 13

y = -1.29 + 0.36*13 = 3.36

The calculations are:

x	y	(x - xbar)	(y - ybar)	(x - xbar)*(y - ybar)	(x - xbar)²	(y - ybar)²
10	2	-2	-1	2	4	1
11	3	-1	0	0	1	0
12	3	0	0	0	0	0
15	4	3	1	3	9	1
xbar (Average)	ybar (Average)	Σ(x - xbar)	Σ(y - ybar)	Σ(x - xbar)*(y - ybar)	Σ(x - xbar)²	Σ(y - ybar)²
12	3	0	0	5	14	2
r = Σ(x - xbar)*(y - ybar)/√Σ(x - xbar)²*Σ(y - ybar)²	0.944911
b1 = Σ(x - xbar)*(y - ybar)/Σ(x - xbar)²	0.357143
bo = ybar - xbar*b1	-1.28571

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