Question

Body Temperatures For the matched pairs listed in Exercise 1, identify the following components used in the Wilcoxon signed-ranks test:

a. Differences d

b. The ranks corresponding to the nonzero values of | d |

c. The signed-ranks

d. The sum of the positive ranks and the sum of the absolute values of the negative ranks

e. The value of the test statistic T

f. The critical value of T (assuming a 0.05 significance level in a test of no difference between body temperatures at 8 AM and 12 AM)

Short answer

a.The following table shows the differences:

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6

b.The ranks are tabulated below:

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6
Ranks of |d|	5	6	1	\( \times \)	3	4	2

c.The signed-ranks are given below:

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6
Ranks of |d|	5	6	1	\( \times \)	3	4	2
Signed-Ranks	+5	-6	-1	\( \times \)	-3	-4	+2

d.The sum of the positive ranks is equal to 7, and the sum of the negative ranks is equal to 14.

e. The value of the test statistic T is equal to 7.

f. The critical value of T at \(\alpha \) equal to 0.05 is equal to 2.

Step-by-step solution

Given information

Samples are given showing the body temperatures at 8 AM and 12 AM.

Determine the Wilcoxon signed-rank test

The Wilcoxon signed-rank test is used when the researcher wants to study the matched samples orrelated samples. This test belongs to the non-parametric category.

Calculate the differences

a.

The Wilcoxon signed-ranks test is a type of non-parametric test that can be used to test the claim of no difference in the temperatures recorded at 8 AM and 12 AM.

The differences between the values of the two samples are computed by subtracting the value from the second sample from the corresponding value of the first sample:

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6

Determine the ranks

b.

Compute the ranks of the differences by ignoring the signs and sorting the values from lowest to highest.

The lowest value gets a rank equal to 1, and subsequently, the ranks get increased by one until the maximum value.

Also, discard the values with a difference of 0.

The following table shows the ranks of the differences:

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6
Ranks of |d|	5	6	1	\( \times \)	3	4	2

Determine the signed-ranks

c.

Assign the signs to the ranks corresponding to the sign of the difference.

Temperature (°F) at 8 AM	98	97.6	97.2	98	97	98	98.2
Temperature (°F) at 12 AM	97	98.8	97.6	98	97.7	98.8	97.6
Differences	1	-1.2	-0.4	0	-0.7	-0.8	0.6
Ranks of |d|	5	6	1	\( \times \)	3	4	2
Signed-Ranks	+5	-6	-1	\( \times \)	-3	-4	+2

Determine the sum of the ranks

d.

The sum of the positive ranks is equal to:

\(5 + 2 = 7\)

Thus, the sum of the positive ranks is equal to 7.

The sum of the negative ranks is equal to:

\(\left( { - 6} \right) + \left( { - 1} \right) + \left( { - 3} \right) + \left( { - 4} \right) = - 14\)

Thus, the absolute value of the sum of the negative ranks is equal to 14.

Determine the test statistic

e.

The smaller value of the two sums is considered the value of the test statistic (T).

Thus, T is equal to 7.

Determine the critical Value

f.

Referring to Table A-8, the critical value of T corresponding to n = 7 and\(\alpha = 0.05\)for a two-tailed test is equal to 2.

Thus, the critical value is 2.