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10.5: Matched or Paired Samples

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, \(\mu_{d}\), 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:

\[t = \frac{\bar{x}_{d} - \mu_{d}}{\left(\frac{s_{d}}{\sqrt{n}}\right)}\]

Example 10.5.1

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table. 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 hypnotism? Test at a 5% 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

 Answer

Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "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: \(\bar{x}_{d} = -3.13\) and \(s_{d} = 2.91\) Verify these values.

Let \(\mu_{d}\) be the population mean for the differences. We use the subscript dd to denote "differences."

Random variable: \(\bar{X}_{d} =\) the mean difference of the sensory measurements

\(H_{0}: \mu_{d} \geq 0\)

The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. \(\mu_{d}\) is the population mean of the differences.)

\(H_{a}: \mu_{d} < 0\)

The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, 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 \(t_{7}\). (Notice that the test is for a single population mean.)

Calculate the p-value using the Student's-t distribution: \(p\text{-value} = 0.0095\)

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 10.5.1.

\(\bar{X}_{d}\) is the random variable for the differences.

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

\(\bar{x}_{d} = -3.13\)

\(s_{d} = 2.91\)

Compare \(\alpha\) and the \(p\text{-value}\): \(\alpha = 0.05\) and \(p\text{-value} = 0.0095\). \(\alpha > p\text{-value}\).

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\). This means that \(\mu_{d} < 0\) and there is improvement.

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.

Note 10.5.1.1

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.

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 \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq:. Arrow down to \(\mu\): and arrow over to < \(\mu_{0}\). Press ENTER. Arrow down to Calculate and press ENTER. The \(p\text{-value}\) is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER.

Exercise 10.5.1

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% 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

Answer

The \(p\text{-value}\) is 0.0130, so we can reject the null hypothesis. There is enough evidence to suggest that the diet lowers cholesterol.

Example 10.5.2

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

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.

\[\bar{x}_{d} = 21.3\), \(s_{d} = 46.7\]

Note 10.5.2.1

The data given here would indicate that the distribution is actually 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 actually negative.

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

Define the random variable: \(\bar{X}\) mean difference in the maximum lift per player.

The distribution for the hypothesis test is \(t_{3}\).

\(H_{0}: \mu_{d} \leq 0\), H_{a}: \mu_{d} > 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 10.5.2.

Calculate the \(p\text{-value}\): The \(p\text{-value}\) is 0.2150

Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because \(\alpha < p\text{-value}\).

What is the conclusion?

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.

Exercise 10.5.2

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 recorded in Table. Are the scores, on average, higher after the class? Test at a 5% 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

Answer

The \(p\text{-value}\) is 0.0874, so we decline to reject the null hypothesis. The data do not support that the class improves SAT scores significantly.

Example 10.5.3

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 data were collected and recorded in Table.

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.

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. \(\bar{x} = 3.71\), \(s_{d} = 4.5\).

Random variable: \(\bar{X} =\) mean difference in the distances between the hands.

Distribution for the hypothesis test: \(t_{6}\)

\(H_{0}: \mu_{d} = 0 H_{a}: \mu_{d} \neq 0\)

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 10.5.3.

Calculate the p-value: The \(p\text{-value}\) is 0.0716 (using the data directly).

(test statistic = 2.18. \(p\text{-value} = 0.0719\) using \((\bar{x}_{d} = 3.71, s_{d} = 4.5\).

Decision: Assume \(\alpha = 0.05\). Since \(\alpha < p\text{-value}\), Do not reject \(H_{0}\).

Conclusion: At the 5% 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.

Exercise 10.5.3

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. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% 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

Answer

The \(p\text{-level}\) is 0.0230, so we can reject the null hypothesis. The data show that the players do not throw the same distance with their off-hands as they do with their dominant hands.

Chapter Review

A hypothesis test for matched or paired samples (\(t\)-test) has these characteristics:

  • Test the differences by subtracting one measurement from the other measurement
  • Random Variable: \(x_{d} =\) mean of the differences
  • Distribution: Student’s-t distribution with \(n - 1\) degrees of freedom
  • If the number of differences is small (less than 30), the differences must follow a normal distribution.
  • Two samples are drawn from the same set of objects.
  • Samples are dependent.

Formula Review

Test Statistic (t-score): \[t = \frac{\bar{x}_{d}}{\left(\frac{s_{d}}{\sqrt{n}}\right)}\]

where:

\(x_{d}\) is the mean of the sample differences. \(\mu_{d}\) is the mean of the population differences. \(s_{d}\) is the sample standard deviation of the differences. \(n\) is the sample size.

Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.

Installation A B C D E F G H
Before 3 6 4 2 5 8 2 6
After 1 5 2 0 1 0 2 2

Exercise 10.5.4

What is the random variable?

Answer

the mean difference of the system failures

Exercise 10.5.5

State the null and alternative hypotheses.

Exercise 10.5.6

What is the  \(p\text{-value}\)?

Answer

0.0067

Exercise 10.5.7

Draw the graph of the \(p\text{-value}\).

Exercise 10.5.8

What conclusion can you draw about the software patch?

Answer

With a \(p\text{-value} 0.0067\), we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level.

Subject A B C D E F
Before 3 4 3 2 4 5
After 4 5 6 4 5 7

Exercise 10.5.9

State the null and alternative hypotheses.

Exercise 10.5.10

What is the \(p\text{-value}\)?

Answer

0.0021

Exercise 10.5.11

What is the sample mean difference?

Exercise 10.5.12

Draw the graph of the \(p\text{-value}\).

Answer

This is a normal distribution curve with mean equal to zero. The values 0 and 1.67 are labeled on the horiztonal axis. A vertical line extends from 1.67 to the curve. The region under the curve to the right of the line is shaded to represent p-value = 0.0021.

Figure 10.5.4.

Exercise 10.5.13

What conclusion can you draw about the juggling class?

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.

Patient A B C D E F
Before 161 162 165 162 166 171
After 158 159 166 160 167 169

Exercise 10.5.14

State the null and alternative hypotheses.

Answer

\(H_{0}: \mu_{d} \geq 0\)

\(H_{a}: \mu_{d} < 0\)

Exercise 10.5.15

What is the test statistic?

Exercise 10.5.16

What is the \(p\text{-value}\)?

Answer

0.0699

Exercise 10.5.17

What is the sample mean difference?

Exercise 10.5.18

What is the conclusion?

Answer

We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective.

Bringing It Together

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test.

  1. independent group means, population standard deviations and/or variances known
  2. independent group means, population standard deviations and/or variances unknown
  3. matched or paired samples
  4. single mean
  5. two proportions
  6. single proportion

Exercise 10.5.19

A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet.

Exercise 10.5.20

A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

Answer

e

Exercise 10.5.21

The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females.

Exercise 10.5.22

A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.

Answer

d

Exercise 10.5.23

A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively.

Exercise 10.5.24

According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this.

Answer

f

Exercise 10.5.25

According to a recent study, U.S. companies have a mean maternity-leave of six weeks.

Exercise 10.5.26

A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally.

Answer

e

Exercise 10.5.27

A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected:

Pre-course score Post-course score
1 300
960 920
1010 1100
840 880
1100 1070
1250 1320
860 860
1330 1370
790 770
990 1040
1110 1200
740 850

Exercise 10.5.28

University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked.

Answer

f

Exercise 10.5.29

Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown Table. Determine the appropriate test and best distribution to use for that test.

  Left-handed Right-handed
Sample size 41 41
Sample mean 97.5 98.1
Sample standard deviation 17.5 19.2
  1. Two independent means, normal distribution
  2. Two independent means, Student’s-t distribution
  3. Matched or paired samples, Student’s-t distribution
  4. Two population proportions, normal distribution

Exercise 10.5.30

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table.

  Player 1 Player 2 Player 3 Player 4
Mean score before class 83 78 93 87
Mean score after class 80 80 86 86

This is:

  1. a test of two independent means.
  2. a test of two proportions.
  3. a test of a single mean.
  4. a test of a single proportion.

Answer

a