<< Chapter < Page Chapter >> Page >
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.
  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. 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. Thepopulation mean for the differences, μ d , is then tested using a Student-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 )

Matched or paired samples

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results forrandomly 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: x d = -3.13 and s d = 2.91 Verify these values.

Let μ d be the population mean for the differences. We use the subscript d to denote "differences."

Random Variable: X d = the mean difference of the sensory measurements

H o : μ d 0
There is no improvement. ( μ d is the population mean of the differences.)

H a : μ d < 0
There is improvement. The score should be lower after hypnotism so the difference ought to be negativeto indicate improvement.

Distribution for the test: The distribution is a student-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-t distribution: p-value = 0.0095


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.

X d is the random variable for thedifferences.

The sample mean and sample standarddeviation of the differences are:

x d = -3.13

s 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 H o .

This means that μ 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 afterhypnotism. 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 calculatorwill 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 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 , the name of the list where you put the data, and 1 for Freq:. Arrow down to μ : and arrow over to < 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 exceptarrow to Draw (instead of Calculate ). Press ENTER .
Got questions? Get instant answers now!

A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the benchpress 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 developmentclass. 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. Thedata 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.

x d = 21.3 s d = 46.7

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

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

The distribution for the hypothesis test is t 3 .

H o : μ d 0 H a : μ d > 0


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.

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 α < p-value .

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Seven eighth graders at Kennedy Middle School measured how farthey 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 eitherhand. 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?

H 0 : μ d equals 0; H a : μ d does not equal 0; Do not reject the null; At a 5% significance level, from the sample data, there is not sufficient evidence to conclude that the mean difference in distances between the children's dominant versus weaker hands is significant (there is not sufficient evidence to show that the children could push the shot-put further with their dominant hand). Alpha and the p-value are close so the test is not strong.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Questions & Answers

Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
how to fabricate graphene ink ?
for screen printed electrodes ?
What is lattice structure?
s. Reply
of graphene you mean?
or in general
in general
Graphene has a hexagonal structure
On having this app for quite a bit time, Haven't realised there's a chat room in it.
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.
Nerisha Reply

Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Collaborative statistics' conversation and receive update notifications?