Question

A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated to establish an appropriate treatment program and estimated length of stay. To see if the evaluation teams are consistent, 12 randomly chosen patients are separately evaluated by two expert teams (A and B) as shown.

  

Estimated Length of Stay in Weeks
Patient
Team	1	2	3	4	5	6	7	8	9	10	11	12
A	38	27	41	46	49	29	46	33	20	13	49	15
B	51	47	40	23	20	16	50	43	52	29	22	21
(a) Choose the appropriate hypotheses. Assume ?d is the difference in mean lengths of team A and team B. H0: ?d ? 0 vs. H1: ?d > 0 H0: ?d = 0 vs. H1: ?d ? 0 H0: ?d ? 0 vs. H1: ?d < 0 (b) Specify the decision rule at the .10 level of significance.(Round your answers to 3 decimal places. A negative values should be indicated by a minus sign.) Reject the null hypothesis if the p-value is (Click to select)greater than or less than 0.10 or if tcalc <  or tcalc > . (c) Find the test statistic tcalc. (Round your answer to 3 decimal places. A negative value should be indicated by a minus sign.) tcalc             (d) Find the p-value. (Round your answer to 3 decimal places. A negative value should be indicated by a minus sign.) p-value             (e) Make a decision. We (Click to select)Do not reject or Reject the null hypothesis. (f) State your conclusion. We (Click to select)can or cannot conclude that the evaluator teams are consistent in their estimates.

Answer 1

(a)

Correct option:

H0: = 0 vs Ha: 0

(b)

= 0.10

ndf = n - 1 = 12 - 1 = 11

From Table critical values of t = 1.7959

Rejection rule:

Reject the null hypothesis if p-value is less than 0.10

OR

if

t_calc < - 1.7959

or

t_calc > 1.7959

(c)

Team A (x)              Team B (y)                   d = x - y

38                                 51                          - 13

27                                  47                         - 20

41                                 40                          1

46                                  23                          23

49                                   20                          29

29                                  16                           13

46                                  50                           - 4

33                                  43                          - 10

20                                  52                           - 32

13                                  29                           - 16

49                                  22                           - 27

15                                  21                          - 6

From the d values, the following statistics are calculated:

n = 12

= - 62/12 = - 5.1667

s_d = 18.9489

SE = s_d/

= 18.9489/ = 5.4701

Test statistic is:

t = /SE

= - 5.1667/5.4701 = - 0.945

(d)

t score = - 0.945

ndf = 12 - 1 = 11

= 0.10

By Technology, p-value = 0.365

(e) Since p-value is greater than , Fail to reject H0.

(f) We can conclude that the evaluator teams are consistent in their estimates.