MATCHED PAIRS t PROCEDURE (AN EXAMPLE)



AP Statistics: Chapter 25 Name______________________________

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 was 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 |(B): # plates correctly identified |DIFFERENCE |

|after training. |before training. |(A) - (B) |

|6 |6 | |

|8 |5 | |

|6 |6 | |

|7 |5 | |

|9 |7 | |

|8 |5 | |

|9 |4 | |

|6 |6 | |

|7 |7 | |

|5 |8 | |

|9 |4 | |

|8 |5 | |

|6 |4 | |

|8 |6 | |

|6 |7 | |

|Mean of DIFFERENCE column | |

|s for DIFFERENCE column | |

MATCHED PAIRS t PROCEDURE (AN EXAMPLE) (Key)

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 was 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 |(B): # plates correctly identified |DIFFERENCE |

|after training. |before training. |(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 |

KEY:

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

Ho: μD = 0 (There is no mean difference/improvement in ability to identify plates)

Ha: μD > 0 (There is a positive mean difference/improvement in ability to identify plates)

Assumptions/Conditions:

SRS ( Given

n = 15 (medium, must check distribution of sample)

The distribution of the sample data is roughly symmetric and unimodal, so t-procedure is ok.

10%: There are over 150 policemen in the force.

The two sets of data (after and before the training) are dependent because each pair came from the same policeman.

Sample mean = 1.5333.

Sample s = 2.1996.

Name: Type of test: paired t-test for the mean difference/improvement.

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.

p-value = P(x-bar > 1.5333) = P(t14 > 2.70) = .0086

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

Conclusion: Our sample t is in the critical region and our p-value is less than 𝜶 of 5%. 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. If there is no improvement in ability to identify plates before and after the training program, then we expect to see a sample mean difference of 1.5333 only .86% of the time due to chance alone.

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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.

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