Question

Compute both t-tests on excel type a conclusion for each problem

Mr. V. wants to know if grades for his classes have changed since they went online. To test for it, he decided to compute an independent samples t-test and a related samples t-test.

1) For an independent sample, he compared grades for his class on social psychology with grades for his class on abnormal psychology. These were the results:

Social psychology	Abnormal psychology
89	75
91	57
86	64
90	55
100	74
98	63
78	87
84	81
89	53
93	40

2) For a related sample, he compared grades for his introduction to psychology class before and after they went online.

Students	Before online	After online
1	78	93
2	85	86
3	56	88
4	74	89
5	43	96
6	67	91
7	47	87
8	58	99
9	64	79
10	70	83

Answer 1

1.

Suppose, random variables X and Y denote grades in social psychology and abnormal psychology respectively.

Here, two different groups of students are used to collect data from two different classes. Further we do not know population standard deviation (or variance). So, we have to perform two sample t-test.

We have to test for null hypothesis

Ho = Hy

against the alternative hypothesis $\tiny H_1:\mu_x\neq\mu_y$

Corresponding calculations in EXCEL are as follows.

D Description E Values F Formula used (start with '=') А B |c Social Abnormal 1 psychology psychology 2 89 75 3 91 57 4 86 64 5 90 55 6 100 74 7 98 63 8 78 87 9 84 81 10 89 53 11 93 40 12 13 14 First sample size 10.00000 COUNT(A2:A11) Second sample size 10.00000 COUNT(B2:B11) First sample mean 89.80000 AVERAGE(A2:A11) Second sample mean 64.90000 AVERAGE(B2:311) First sample variance 41.28889 VAR(A2:A11) Second sample variance 206.54444 VAR(B2:311) Difference in means 24.90000 E4-E5 Pooled variance 123.91667 ((E2-1)*E6+(E3-1)"E7)/(E2+E3-2) Standard error 4.97829 SQRT(E9*(1/E2+1/E3)) Test statistic (t) value 5.00172 E8/E10 Degrees of freedom 18.00000 E2+E3-2 Number of tails 2.00000 as not equal to type alternative hypothesis P-value 0.00009 TDIST(E11,E12,2)

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$ , level of significance.

We generally test for level of significance 0.10, 0.05, 0.01 or something like these.

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that grades are different in social psychology and abnormal psychology.

2.

Here, in two different instances (before and after online classes) grades were observed for same set of students. So, we have to perform paired t-test.

Suppose, random variables X and Y denote grades before and after online classes respectively. Also, random variable D(=Y-X) denotes difference in grades.

We have to test for null hypothesis $\tiny H_0:\bar d=0$

against the alternative hypothesis $\tiny H_1:\bar d\neq 0$

Corresponding calculations in EXCEL are as follows.

D с D=Y-X E Description F Values G Formula used (start with '=') 56 А B Before After 1 online (1) online (Y) 2 78 93 3 85 86 4 88 5 74 89 6 43 96 7 67 91 8 47 9 58 99 10 64 79 11 70 83 15 1 32 15 53 24 40 41 15 Sample size 10.00000 COUNT(C2:021) Mean 24.90000 AVERAGE(C2:C21) Standad deviation 16.10693 STDEV(C2:C21) Standard error 5.09346 F4/SQRT(F2) Test statistic (t) value 4.88862 F3/F5 Degrees of freedom 9.00000 F2-1 Number of tails 2.00000 as not equal to type alternative hypothesis P-value 0.00086 TDIST(F6,F7,F8) 87 13

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$ , level of significance.

We generally test for level of significance 0.10, 0.05, 0.01 or something like these.

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that grades are different before and after online classes.