Question

An educator believes that new reading activities for elementary school children will improve reading comprehension scores. She randomly assigns third graders to an eight-week program in which some will use these activities and others will experience traditional teaching methods. At the end of the experiment, both groups take a reading comprehension exam. Their Scores are shown below.

New Activities: 23, 34, 34, 34, 35, 37, 42, 43, 44, 46, 47, 48, 49, 55

Control: 12, 13, 22, 26, 29, 30, 32, 32, 32, 32, 32 34, 36, 43, 45, 45, 64

**This data was given in a stem and leaf display

Test an Appropriate hypothesis at the state your conclusion. Use a 5% significance level or $\alpha$ =0.05.

t-statistic, t=2.13

The P-value is 0.021

1) Use the Wilcoxon rank sum test to test an appropriate hypothesis. Compute​ W, the test​ statistic, for the New Activities group.

W=?
2) Find the P-value.

The P-value is ?

**I actually have the answer to this but cannot wrap my head around how they got the answer. I ran into that troube to regarding the dataset not being paired but i literally am not given anything else.

Answer: W=288 and the P-value=0.006

HELP!

Answer 1

1.) As per Question, the Hypotheses becomes:

Ho: There is no change in average score of students who attended the program and who doesn't.

Mx = My

Ha: There is change in average score of students who attended the program and who doesn't.

Mx $\neq$ My

Given data for New Activities:

New(X)
23
34
34
34
35
37
42
43
44
46
47
48
49
55

Mean, M_x = 40.7857

New(X)	Diff (X - Mx)	Sq. Diff (X - Mx)2
23	-17.786	316.332
34	-6.786	46.046
34	-6.786	46.046
34	-6.786	46.046
35	-5.786	33.474
37	-3.786	14.332
42	1.214	1.474
43	2.214	4.903
44	3.214	10.332
46	5.214	27.189
47	6.214	38.617
48	7.214	52.046
49	8.214	67.474
55	14.214	202.046

N_x: 14
df_x = N - 1 = 14 - 1 = 13
M_x: 40.79
Sum Of Square(SS_x): 906.357
s²_x = SS_x/(N_x - 1) = 906.357/(14-1) = 69.72

Given Control Data:

Control(Y)
12
13
22
26
29
30
32
32
32
32
32
34
36
43
45
45
64

M_y = 32.8823

Control(Y)	Diff (Y - My)	Sq. Diff (Y - My)2
12	-20.882	436.073
13	-19.882	395.308
22	-10.882	118.426
26	-6.882	47.367
29	-3.882	15.073
30	-2.882	8.308
32	-0.882	0.779
32	-0.882	0.779
32	-0.882	0.779
32	-0.882	0.779
32	-0.882	0.779
34	1.118	1.249
36	3.118	9.720
43	10.118	102.367
45	12.118	146.837
45	12.118	146.837
64	31.118	968.308

N_y: 17
df_y = N - 1 = 17 - 1 = 16
M_y: 32.8823
SS_y: 2399.765
s²_y = SS_y/(N_y - 1) = 2399.765/(17-1) = 149.99

Formula for t-statistic is:

$t_independent_means.png$

T-value Calculation

s²_p = ((df_x/(df_x + df_y)) * s²_x) + ((df_y/(df_x + df_y)) * s²_y)

s²_p = ((13/29) * 69.72) + ((16/29) * 149.99) = 114

s²_Mx = s²_p/N_x = 114/14 = 8.1433
s²_My = s²_p/N_y = 114/17 = 6.7062

t = (Mx- M_y)/√(s²_Mx + s²_My) = 7.9033/√14.85 = 2.05

So, t = 2.05

Hence corresponding p -value is, p = 0.0247

Since, given alpha = 0.05

P < 0.05, so we can reject the Null Hypotheses(Ho)

The alternate Hypotheses(Ha) is accepted.

2.)

For Wilcoxon rank sum test, Hypotheses are:

Ho: Median Scores for both the groups is same

MD_x = MD_y

Ha: Median Scores for both the groups is not same

MD_x $\neq$ MD_y

First, we put both samples together and organize it in ascending order, which is shown in the table below:

Values	Sample
12	2
13	2
22	2
23	1
26	2
29	2
30	2
32	2
32	2
32	2
32	2
32	2
34	2
34	1
34	1
34	1
35	1
36	2
37	1
42	1
43	2
43	1
44	1
45	2
45	2
46	1
47	1
48	1
49	1
55	1
64	2

Now, that the values that are in ascending order are assigned ranks to them, taking care of assigning the average rank to values with rank ties

Values	Sample	Rank	Rank (Adjusted for ties)
12	2	1	1
13	2	2	2
22	2	3	3
23	1	4	4
26	2	5	5
29	2	6	6
30	2	7	7
32	2	8	10
32	2	9	10
32	2	10	10
32	2	11	10
32	2	12	10
34	2	13	14.5
34	1	14	14.5
34	1	15	14.5
34	1	16	14.5
35	1	17	17
36	2	18	18
37	1	19	19
42	1	20	20
43	2	21	21.5
43	1	22	21.5
44	1	23	23
45	2	24	24.5
45	2	25	25
46	1	26	26
47	1	27	27
48	1	28	28
49	1	29	29
55	1	30	30
64	2	31	31

Note : For Rank Adjusted with ties, we will take average of corresponding ranks and fill the same

average rank for all values

The sum of adjusted ranks for sample 1 is:

R₁= 4+14.5+14.5+14.5+17+19+20+21.5+23+26+27+28+29+30 = 288

and the sum of ranks of sample 2 is:

R₂ = 1+2+3+5+6+7+10+10+10+10+10+14.5+18+21.5+24.5+24.5+31 = 208

Hence, the test statistic is W = R₁ = 288

To get P value, we need to calulate Z statistic:

The z-statistic is computed as follows:

z = [W - (n1* ​​(n1​+n2​+1) / 2) ] / $\sqrt{[n1 * n2 * (n1 + n2 +1)] / 12}$

z = 2.54

Checking for p value in z score tabe, we get

p = 0.0055 = 0.006

since p < 0.05, we can reject the Null Hypotheses.

So, Alternate Hypotheses is accepted and there is enough evidence for below expression

MD_x $\neq$ MDy