Question

You want to test your hypothesis below:

There is a difference in mean test scores between two classes: Class 1 vs. Class 2

            Provide your answers in the template below the data set.

class 1	class 2
15	15
23	12
22	17
18	17
19	19
18	17
15	17
19	18
16	13
21	20
15	19
14	21
20	21
23	14
19	19
17	17
17	17
21	18
7	13
16	20
14	19
17	21

State the Ha=
State the H0=
	Class 1	Class 2
Mean
SD
t-test, p value
Conclusion (reject or accept the null)
Interpret the conclusion

Answer 1

X1	x2	(x1-x1bar)^2	(x2-x2bar)^2
15 23 22 18 19 18 15 19 16 21 15 14 20 23 19 17 17 21 7 16 14 7	15 12 17 17 19 17 17 18 13 20 19 21 21 14 19 17 17 18 13 20 19 21	4.37 34.92 24.10 0.83 3.64 0.83 4.37 3.64 1.19 15.28 4.37 9.55 8.46 34.92 3.64 0.01 0.01 15.28 101.83 1.19 9.55 101.83 SS: 383.82	6.02 29.75 0.21 0.21 2.39 0.21 0.21 0.30 19.84 6.48 2.39 12.57 12.57 11.93 2.39 0.21 0.21 0.30 19.84 6.48 2.39 12.57 SS: 149.45

For class1

N1: 22
df1 = N - 1 = 22 - 1 = 21
M1: 17.09
SS1: 383.82
s21 = SS1/(N - 1) = 383.82/(22-1) = 18.28


Class 2

N2: 22
df2 = N - 1 = 22 - 1 = 21
M2: 17.45
SS2: 149.45
s22 = SS2/(N - 1) = 149.45/(22-1) = 7.12

Null and alternate hypothesis

H0:

H1:

2) test statistics

s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((21/42) * 18.28) + ((21/42) * 7.12) = 12.7

s2M1 = s2p/N1 = 12.7/22 = 0.58
s2M2 = s2p/N2 = 12.7/22 = 0.58

t = (M1 - M2)/√(s2M1 + s2M2) = -0.36/√1.15 = -0.34

3) p value = 0.7367

4).accept the null hypothesis

Conclusion there insufficient evidence to conclude that there is a difference in mean test scores between two classes: Class 1 vs. Class 2.