Introduction

Introduction

When using a hypothesis test for matched or paired samples, the following characteristics should be present:

  1. Simple random sampling is used.
  2. Sample sizes are often small.
  3. Two measurements (samples) are drawn from the same pair of individuals or objects.
  4. Differences are calculated from the matched or paired samples.
  5. The differences form the sample that is used for the hypothesis test.
  6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that 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, μd, is then tested using a Student’s t-test for a single population mean with n – 1 degrees of freedom, where n is the number of differences.

The test statistic (t-score) is

t=x¯dμd(sdn).t=x¯dμd(sdn).

Example 10.11

A study was conducted to investigate the effectiveness of pain-reducing medication. Results for randomly selected subjects are shown in Table 10.11. A lower score indicates less pain. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after the medication? Test at a 5 percent significance level.

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
Table 10.11
Solution 10.11

Corresponding before and after values form matched pairs. (Calculate afterbefore.)

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
Table 10.12

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: xd¯=–3.13xd=–3.13 and sd=2.91sd=2.91 Verify these values.

Let μdμd be the population mean for the differences. We use the subscript dd to denote differences.

Random variable: X¯dX¯d = the mean difference of the sensory measurements.

H0: μd ≥ 0

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. That means the subject shows no improvement. μd is the population mean of the differences.

Ha: μd < 0

The alternative hypothesis is negative, meaning there is less pain felt after taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, so the difference ought to be negative to indicate improvement.

Distribution for the test: The distribution is a Student’s t with df = n – 1 = 8 – 1 = 7. Use t7. Note that the test is for a single population mean.

Calculate the p-value using the Students t-distribution: p-value = 0.0095

Graph:

Normal distribution curve of the average difference of sensory measurements with values of -3.13 and 0. A vertical upward line extends from -3.13 to the curve, and the p-value is indicated in the area to the left of this value.
Figure 10.10

X¯dX¯d is the random variable for the differences.

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

x¯dx¯d = –3.13

s¯ds¯d = 2.91

Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value.

Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement.

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

Note

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.

Using the TI-83, 83+, 84, 84+ Calculator

Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for μ0μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to < μ0μ0. Press 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.

Try It 10.11

A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5 percent level.

Subject A B C D E F G H I
Before 209 210 205 198 216 217 238 240 222
After 199 207 189 209 217 202 211 223 201
Table 10.13

Example 10.12

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 four 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 weight lifted prior to the class 205 241 338 368
Amount of weight lifted after the class 295 252 330 360
Table 10.14

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}. Assume the differences have a normal distribution.

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

x¯dx¯d = 21.3, sd = 46.7

Note

The data given here would indicate that the distribution is right-skewed. The difference 90 may be an extreme outlier. It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are negative.

Using the difference data, this becomes a test of a single __________.

Define the random variable: X¯dX¯d mean difference in the maximum lift per player.

The distribution for the hypothesis test is t3.

H0: μd ≤ 0, Ha: μd > 0

Graph:

Normal distribution curve with values of 0 and 21.3. A vertical upward line extends from 21.3 to the curve and the p-value is indicated in the area to the right of this value.
Figure 10.11

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

Decision: If the level of significance is 5 percent, the decision is not to reject the null hypothesis, because α < p-value.

What is the conclusion?

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

Try It 10.12

A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data are recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5 percent level.

SAT Scores Student 1 Student 2 Student 3 Student 4
Score before class 1840 1960 1920 2150
Score after class 1920 2160 2200 2100
Table 10.15

Example 10.13

Seven eighth graders at Kennedy Middle School measured how far they could push the shot put with their dominant (writing) hand and their weaker (nonwriting) hand. They thought that they could push equal distances with both hands. The data were collected and recorded in Table 10.16.

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
Table 10.16

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

Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation. x¯dx¯d = 3.71, sdsd = 4.5.

Random variable: X¯dX¯d = mean difference in the distances between the hands.

Distribution for the hypothesis test: t6

H0: μd = 0 Ha: μd ≠ 0

Graph:

This is a normal distribution curve with mean equal to zero. Both the right and left tails of the curve are shaded. Each tail represents 1/2(p-value) = 0.0358.
Figure 10.12

Calculate the p-value: The p-value is 0.0716 (using the data directly).

(Test statistic = 2.18. p-value = 0.0719 using (x¯d=3.71, sd=4.5)..(x¯d=3.71, sd=4.5.)

Decision: Assume α = 0.05. Since α < p-value, do not reject H0.

Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot put.

Try It 10.13

Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean 5 difference in distances between the dominant and off-hand is significant. Test at the 5 percent level.

  Player 1 Player 2 Player 3 Player 4 Player 5
Dominant Hand 120 111 135 140 125
Off-Hand 105 109 98 111 99
Table 10.17