Summary: This module provides an overview of Hypothesis Testing: Matched or Paired Samples as a part of Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

- Simple random sampling is used.
- Sample sizes are often small.
- Two measurements (samples) are drawn from the same pair of individuals or objects.
- Differences are calculated from the matched or paired samples.
- The differences form the sample that is used for the hypothesis test.
- The matched pairs have differences that either come from a population that is normal or the number of differences is sufficiently large so the distribution of the sample mean of differences is approximately normal.

In a hypothesis test for matched or paired samples, subjects are matched in
pairs and differences are calculated. The differences are the data. The
population mean for the differences,

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the table. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution.

Subject: | A | B | C | D | E | F | G | H |
---|---|---|---|---|---|---|---|---|

Before | 6.6 | 6.5 | 9.0 | 10.3 | 11.3 | 8.1 | 6.3 | 11.6 |

After | 6.8 | 2.4 | 7.4 | 8.5 | 8.1 | 6.1 | 3.4 | 2.0 |

Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.

Corresponding "before" and "after" values form matched pairs. (Calculate "sfter" - "before").

After Data | Before Data | Difference |
---|---|---|

6.8 | 6.6 | 0.2 |

2.4 | 6.5 | -4.1 |

7.4 | 9 | -1.6 |

8.5 | 10.3 | -1.8 |

8.1 | 11.3 | -3.2 |

6.1 | 8.1 | -2 |

3.4 | 6.3 | -2.9 |

2 | 11.6 | -9.6 |

The data *for the test* are the
differences:
{0.2, -4.1, -1.6, -1.8, -3.2, -2, -2.9, -9.6}

The sample mean and sample standard
deviation of the differences are:

Let

*Random Variable:*

*Distribution for the test:* The distribution is a student-t with

*Calculate the p-value using the Student-t distribution:*

*Graph:*

The sample mean and sample standard deviation of the differences are:

*Compare *

*Make a decision:* Since

This means that

*Conclusion:* At a 5% level of significance, from the sample data, there is sufficient
evidence to conclude that the sensory measurements, on average, are lower after
hypnotism. Hypnotism appears to be effective in reducing pain.

For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead
of time (*after - before*) and put the differences into a list or you can put the *after* data
into a first list and the *before* data into a second list. Then go to a third list
and arrow up to the name. Enter 1st list name - 2nd list name. The calculator
will do the subtraction and you will have the differences in the third list.

TI-83+ and TI-84: Use your list of differences as the data. Press μ 0 μ 0 , the name of the list where you put
the data, and μ 0 μ 0 . Press

`STAT`

and arrow over to `TESTS`

. Press `2:T-Test`

. Arrow over to `Data`

and press
`ENTER`

. Arrow down and enter `0`

for `1`

for Freq:. Arrow down to `μ`

: and arrow over to `<`

`ENTER`

. Arrow down to `Calculate`

and press `ENTER`

. The p-value is
0.0094 and the test statistic is -3.04. Do these instructions again except
arrow to `Draw`

(instead of `Calculate`

). Press `ENTER`

.
A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked 4 of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:

Weight (in pounds) | Player 1 | Player 2 | Player 3 | Player 4 |
---|---|---|---|---|

Amount of weighted lifted prior to the class | 205 | 241 | 338 | 368 |

Amount of weight lifted after the class | 295 | 252 | 330 | 360 |

*The coach wants to know if the strength development class makes his players
stronger, on average.*

Record the *differences* data. Calculate the differences by subtracting the amount of
weight lifted prior to the class from the weight lifted after completing the class. The
data for the differences are:
{90, 11, -8, -8}. The differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation.

Using the difference data, this becomes a test of a single __________ (fill in the blank).

*Define the random variable:*

The distribution for the hypothesis test is

*Graph:*

*Calculate the p-value:* The p-value is 0.2150

*Decision:* If the level of significance is 5%, the decision is to not reject the null
hypothesis because

*What is the conclusion?*

means; At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.

Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The following data was collected.

Distance (in feet) using | Student 1 | Student 2 | Student 3 | Student 4 | Student 5 | Student 6 | Student 7 |
---|---|---|---|---|---|---|---|

Dominant Hand | 30 | 26 | 34 | 17 | 19 | 26 | 20 |

Weaker Hand | 28 | 14 | 27 | 18 | 17 | 26 | 16 |

*Conduct a hypothesis test* to determine whether the mean difference in distances between the children's
dominant versus weaker hands is significant.

use a t-test on the difference data. Assume the differences have a normal distribution. The random variable is the mean difference.

The test statistic is 2.18 and the p-value is 0.0716.

*What is your conclusion?*

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