Question

A study has been constructed to observe the efficiency of the new method of teaching science subject of a small class. All students of the class are required to take the test before the method of teaching being conducted, and the test after being implemented. The marks are given in Table 2, 4. Table 2: Marks for Science Subject Before 85 After Student 85 42 S2 53 67 59 S3 78 86 S4 S5 S6 S7 62 51 67 56 89 93 S8 S9 62 68 89 86 s10 75 72 S11 S12 71 72 48 45 S13 S14 75 82 69 70 67 S15 50 Find the 90% con fidence interval of the difference scores between before and after implementation. (a) [9 marks] Thus, at 5% significance level, test whether the implementation efficiently increase the students' marks. (b) 13 marks]

Answer 1

(b)

The null hypothesis assumes that the true mean difference between the paired samples is zero

H0 : No effect of implementation on students marks.

H1:- implementation effectively increases students marks.

Sq. Dev Dev (Diff- M) Diff (T2- T1) Treatment 2 Treatment 1 2.56 1.6 85, 42. 59, 86. 51, 85. 53, 67, 78 62, 88.36 -9.4 -11 40.96 -6.4 -8 92.16 9.6 8 88.36 -9.4 -11 88.36 -9.4 -11 56, 67, 89. 31.36 5.6 4 93. 57,76 7.6 6 68 89, 62. 21.16 4.6 3 1.96 86. -1.4 -3 72, 75. 71. 48, 69, 6.76 2.6 1 72, 45, 1.96 -1.4 -3 57.76 7.6 6 75, 82, s0 184.96 13.6 12 70. 67 237.16 -15.4 -17 S: 1001.6 M: -1.6 m m

Mean: -1.6
μ = 0
S² = SS⁄df = 1001.6/(15-1) = 71.54
S²_M = S²/N = 71.54/15 = 4.77
S_M = √S²_M = √4.77 = 2.18

T-value Calculation

t = (M - μ)/S_M = (-1.6 - 0)/2.18 = -0.73

The value of t is -0.732626. The value of p is .23794. The result is not significant at p < .05

If the p-value is greater to the alpha (p< .05), then we fair to reject the null hypothesis

So No effect of implementation on students marks.

(a)

from above, s²= 71.54

So s= 8.46

df=n-1=15-1=14

alpha = 1-confidence% = 1-0.9 = 0.1

So spread range is alpha/2 = 0.1 / 2 = 0.05

T- score at 0.05 = 2.1448

Calculate high end confidence interval total: High End dtscorea X S/Nn + 1.62.1448 x 8.4583010789908/15 High End + High End 1.6 2.1448 x 8.4583010789908/3.8729833462074 High End 1.6 + 4.6840800831184 High End 6.2841 Calculate low end confidence interval total: Low End d - tscorea X SdVn Low End 1.6 2.1448 x 8.4583010789908/115 Low End 1.6 - 2.1448 x 8.4583010789908/3.8729833462074 Low End 1.6 + 4.6840800831184 Low End -3.0841

So 90% confidence interval is

-3.0841 < d-bar < 6.2841