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 | 46 | 48 | 42 | 32 | 20 | 31 | 35 | 47 | 25 | 17 | 37 | 46 |
B | 38 | 34 | 17 | 45 | 35 | 21 | 50 | 31 | 17 | 43 | 25 | 20 |
Find the test statistic tcalc. (Use
Minitab. Round your answer to 4 decimal places. A
negative value should be indicated by a minus
sign.)
tcalc
Find the p-value. (Use Minitab. Round your
answer to 4 decimal places.)
p-value
This is a two-tailed test. Using the given data to solve using Excel:
Calculate the Difference (dbar) column by subtracting the B from A. Find the mean difference and standard deviation of the difference column to find the test statistics and p-value.
Team | A | B | Difference (dbar) |
1 | 46 | 38 | 8 |
2 | 48 | 34 | 14 |
3 | 42 | 17 | 25 |
4 | 32 | 45 | -13 |
5 | 20 | 35 | -15 |
6 | 31 | 21 | 10 |
7 | 35 | 50 | -15 |
8 | 47 | 31 | 16 |
9 | 25 | 17 | 8 |
10 | 17 | 43 | -26 |
11 | 37 | 25 | 12 |
12 | 46 | 20 | 26 |
Total | 50 | ||
Mean Difference | 4.166666667 | ||
Standard Deviation | 17.08712079 | ||
n | 12 | ||
Test statistic | 0.8447 | ||
P-value | 0.4163 |
Test statistics, tcalc = Mean difference/Standard deviation/sqrt(12)
P-value is found by looking in the t-table for test statistic = 0.8447 with the degree of freedom, df = 11.
Therefore, the test statistic, tcalc is 0.8447.
P-value is 0.4163.
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