Sanderson M. Smith

MATCHED PAIRS t PROCEDURE (AN EXAMPLE)

Police trainees were seated in a darkened room facing a projector screen. Ten different license planes were projected on the screen, one at a time, for 5 seconds each, separated by 15-second intervals.

After the last 15-second interval, the lights were turned on and the police trainees were asked to write down as many of the 10 license plate numbers as possible, in any order at all.

A random sample of 15 trainees who took this test where then given a week-long memory training course. They were then retested. The results are shown in the table below.

Test, at the 5% level of significance, that the memory course improved the ability of the trainees to correctly identify license plates.

 (A): # plates correctly identified after training. (B): # plates correctly identified before training. DIFFERENCE (A) - (B) 6 6 0 8 5 3 6 6 0 7 5 2 9 7 2 8 5 3 9 4 5 6 6 0 7 7 0 5 8 -3 9 4 5 8 5 3 6 4 2 8 6 2 6 7 -1 Mean of DIFFERENCE column----> 1.5333 s for DIFFERENCE column--------> 2.1996

We will run a one-sample t test on the DIFFERENCE column. In this situation, m is the mean improvement that would be achieved if the entire population of police trainees took the memory training course.

Ho: m = 0 (There is no mean improvement in ability to identify plates).

Ha: m > 0 (There is mean improvement in ability to identify plates).

Sample mean = 1.5333.
Sample s = 2.1996.
Type of test: One-tail t test.

Standard error of sample mean = 2.1996/SQRT(15) = 0.5679.
Degrees of freedom = 15 - 1 = 14.

Calculated t statistic: t = (1.5333 - 0)/0.5679 = 2.70.

Critical values of t (5%, one-tail, 14 df): t > 1.761

Conclusion: Our sample t is in the critical region. We reject Ho at the 5% level of significance. There is evidence to suggest that the training will result in an improvement in the ability to identify plates.

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Alternate approaches:

(1) The 90% confidence interval for x(bar) = 1.5333 is

1.5333 plus/minus (1.761)(0.5679) = 1.5333 plus/minus 1.000 = [0.5333, 2.533]

This interval does not contain 0. Hence we would reject Ho at the 5% level of significance.

(2) Using the TI-83 calculator, if Ho is true, the probability of getting a t statistic as large as 2.70 is

tcdf(2.70,1E99,14) = 0.0086 = 0.86%. We would reject Ho at the 5% level of significance.

Note: We would also reject Ho at the 1% level of significance, but not at the 1/2% level.