When using a hypothesis test for matched or paired samples, the following characteristics should be present:
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 ¯ d − μ d ( s d n ) t = x ¯ d − μ d ( s d n )A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.14. 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 |
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:
The sample mean and sample standard deviation of the differences are: x – d = –3.13 x – d = –3.13 and s d = 2.91 s d = 2.91 Verify these values.
Let μ d μ d be the population mean for the differences. We use the subscript d d to denote "differences."
Random variable: X ¯ d X ¯ d = the mean difference of the sensory measurements
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. μd is the population mean of the differences.)
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 t7. (Notice that the test is for a single population mean.)
Calculate the test statistic and p-value using the Student's t-distribution:
test statistic, t = –3.036, p-value = 0.0095
Graph:
X ¯ d X ¯ d is the random variable for the differences.
The sample mean and sample standard deviation of the differences are:
Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value.
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 1 st list name - 2 nd 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 μ 0 μ 0 , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ : and arrow over to < μ 0 μ 0 . 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 .
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 |
A college softball coach was interested in whether the college's strength development class increased their players' maximum lift (in pounds). Four players participated in the 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 the 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: . Assume the differences have a normal distribution.
Using the differences data, calculate the sample mean and the sample standard deviation.
x ¯ d = 21.3 , s d = 46.7 x ¯ d = 21.3 , s d = 46.7
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: X ¯ d X ¯ d mean difference in the maximum lift per player.
The distribution for the hypothesis test is t3.
Graph:
Calculate the p-value: The p-value is 0.2150
Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p-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. It should be noted that this data set consists of a very small sample size, with the possibility of an outlier data value. It is important to confirm that the matched pairs have differences that come from a population that is normal. The coach may want to consider increasing the sample size of players in the study.
A new prep class was designed to improve SAT test scores. Four 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 10.18. 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 |
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 10.19.
| 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: . The differences have a normal distribution.
Using the differences data, calculate the sample mean and the sample standard deviation. x ¯ d x ¯ d = 3.71, s d s d = 4.5.
Random variable: X ¯ d X ¯ d = mean difference in the distances between the hands.
Distribution for the hypothesis test: t6
Graph:
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 , s d = 4.5. ) ( x ¯ d = 3.71 , s d = 4.5. )
Decision: Assume α = 0.05. Since α < p-value, Do not reject H0.
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.
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 10.20. 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 |
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