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Wilcoxon signed-ranks test for method Services data of self-reported and measured his for d evidence to support the claim tha
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Answer #1
Reported Height 68 71 63 70 71 60 65 64 54 63 66 72
Measured Height 67.9 69.9 64.9 68.3 70.3 60.6 64.5

67.0

55.6 74.2 65 70.8
Difference(d) 0.1 1.1 -1.9 1.7 0.7 -0.6 0.5 -3 -1.6 -11.2 1 1.2
Rank(d) 1 6 10 9 4 3 2 11 8 12 5 7
Signed Rank +1 +6 -10 +9 +4 -3 +2 -11 -8 -12 +5 +7

a)

Null and Alternative Hypotheses:

H_0:There is no difference in median between self reported heights and measured heights .

H_1:There is a difference in median between self reported heights and measured heights .

b)

Here n=12

The sum of positive ranks is:

w +=1+2+4+5+6+7+9 = 34\

and the sum of negative ranks is:

W-= 3+8 + 10 + 11 + 12 = 44

c)

The test statistic is T = min{W+,w-} = min{34, 44} = 34

d)

The critical value for the significance level g= 0,05 provided, and the type of tail specified is T* = 13 , and the null hypothesis is rejected if T< 13 .

e)

Since in this case T = 34 > 13 , there is not enough evidence to claim that the population median of differences is different than 0, at theg= 0,05 significance level.

Thus we can conclude that there is a significant difference in median between self reported heights and measured heights.

3) Wilcoxon Rank-Sum Test

Male Rank Female Rank
522 21 264 15
127 9 181 12
740 25 267 17
49 2 384 19
230 13 98 6
316 18 62 3
590 24 126 8
466 20 89 5
121 7 531 22
578 23 130 10
78 4 175 11
265 16 44 1
250 14

n_1=13

R_1=196

n_2=12

R_2=129

The sum of ranks for sample 1(Male) is:

Ri=2+4+7+9+13+14+16+18+ 20 +21+23+24+25 = 196

and the sum of ranks of sample 2(Female) is:

R2=1+3+5+6+8+10 + 11 + 12 + 15 + 17 + 19+22 = 129

Hence, the test statistic is  R= R1 = 196.

The following null and alternative hypotheses need to be tested:

H_0:The two populations of men and woman have same median cholesterol levels .or H_0​ : Median (Difference) = 0

H_1:The two populations of men and woman have different median cholesterol levels .or H_1 ​: Median (Difference) \neq 0

Observe that both sample sizes are greater than 10, then we can use normal approximation. The following z-statistic will be used.

The z-statistic is computed as follows:

R-nini+n2+1)/2 ning(nı + n2+1)/(12) 196 – 13(13+12+1)/2) 13 * 12(13+ 12+1)/12 = 1.469

Based on the information provided, the significance level is g= 0,05 , and the critical value for a two-tailed test is .The rejection region for this two-tailed test is R= 2: : > 1.96 .

Since it is observed that 12 = 1.469<%= 1.96 , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1419 , and since p= 0.1419 > 0.05 , it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population median of differences is different than 0, at the 0.05 significance level.

Graphically:

p-value = 0.1419 0.35 0.30 0.25 0.20 0.15- 0.10 0.05 - 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3

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