Question

The grade point average of 4 randomly selected college students is recorded at the end of the fall and spring semesters of their senior year, as follows: Student 1 2 3 4 Fall 2.6 3.0 1.5 1.4 Spring 1.7 2.2 1.4 0.90 Construct a 95% confidence interval for the mean difference in GPA

Answer 1

Answer:

(-0.607 ; 1.757)

Step-by-step explanation:

Student: __ 1 ___ 2 ___ 3 ___ 4

Fall _____2.6__ 3.0__ 1.5 __1.4

Spring___ 1.7__ 2.2 __1.4 __0.90

Using calculator :

Fall:

Xbar1 = 2.125

Sample size, n1 = 4

Standard deviation, s1 = 0.8

Spring :

Xbar2 = 1.55

Sample size, n2 = 4

Standard deviation, s2 = 0.54

Confidence interval :

(xbar1 - xbar2) ± Tcritical * Sp* √1/n1 + 1/n2

Sp = √[(df1*s1² + df2*s2²) ÷ (n1 + n2 - 2)]

df = n - 1

(xbar1 - xbar2) = 2.125 -1.55 = 0.575

Hence,

Sp = √(((3*0.8^2) + (3*0.54^2)) / 6)

Sp= √2.7948 ÷ 6

Sp = √0.4658

Sp= 0.682

Tcritical at 95%, df = 4 = 2.45

Error margin = 2.45 * 0.682 * √1/4 + 1/4 = 1.182

Confidence Interval:

Lower boundary = 0.575 -1.182 = -0.607

Upper boundary = 0.575 + 1.182 = 1.757

(-0.607 ; 1.757)