An education researcher would like to test whether 2nd graders retain or lose knowledge during the summer when they are presumably not in school. She asks nine 2nd graders to take a comprehension exam at the end of the school year (May), and then asks those same students to come back after the summer (late August) to retake a different but equivalent exam, to see if their level of comprehension has changed. Using the data below, test this hypothesis using an (alpha level of 0.05.) May August
95 100
72 80
78 95
50 65
89 85
92 98
75 70
90 96
65 87

Required:
What is the appropirate test?

Kindly check explanation

Step-by-step explanation:

Given the data:

May August

95 100

72 80

78 95

50 65

89 85

92 98

75 70

90 96

65 87

The appropriate test is a paired t test :

d = difference between May and August

d = (-5, -8, -17, -15, 4, -6, 5, -6, -22)

The hypothesis :

H0 : μd = 0

H1 : μd ≠ 0

The test statistic :

T = dbar / (Sdbar/√n)

Where, dbar and Sdbar are the mean and standard deviation of 'd' respectively.

Using calculator :

dbar = - 7.777 ; Sdbar = 9.052

Test statistic = - 7.777 / (9.052 /√9)

Test statistic = - 2.577

The Pvalue, df = n - 1 = 9 - 1 = 8

Pvalue(-2.577, 8) = 0.0327

At α = 0.05

Pvalue < α ; WE reject the H0 ; and conclude that there has been a change in score

the answer is a

step-by-step explanation:

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3