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Statistics: 1.1 Paired t-tests, Study notes of Statistics

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in ...

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Statistics: 1.1 Paired t-tests
Rosie Shier. 2004.
1 Introduction
A paired t-test is used to compare two population means where you have two samples in
which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ diagnostic test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments
where the measurements/treatments are applied to the same subjects (e.g. blood
pressure measurements using a stethoscope and a dynamap).
2 Procedure for carrying out a paired t-test
Suppose a sample of nstudents were given a diagnostic test before studying a particular
module and then again after completing the module. We want to find out if, in general,
our teaching leads to improvements in students’ knowledge/skills (i.e. test scores). We
can use the results from our sample of students to draw conclusions about the impact of
this module in general.
Let x= test score before the module, y= test score after the module
To test the null hypothesis that the true mean difference is zero, the procedure is as
follows:
1. Calculate the difference (di=yixi) between the two observations on each pair,
making sure you distinguish between positive and negative differences.
2. Calculate the mean difference, ¯
d.
3. Calculate the standard deviation of the differences, sd, and use this to calculate the
standard error of the mean difference, SE (¯
d) = sd
n
4. Calculate the t-statistic, which is given by T=¯
d
SE (¯
d). Under the null hypothesis,
this statistic follows a t-distribution with n1 degrees of freedom.
5. Use tables of the t-distribution to compare your value for T to the tn1distribution.
This will give the p-value for the paired t-test.
1
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Statistics: 1.1 Paired t-tests

Rosie Shier. 2004.

1 Introduction

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. Examples of where this might occur are:

  • Before-and-after observations on the same subjects (e.g. students’ diagnostic test results before and after a particular module or course).
  • A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a stethoscope and a dynamap).

2 Procedure for carrying out a paired t-test

Suppose a sample of n students were given a diagnostic test before studying a particular module and then again after completing the module. We want to find out if, in general, our teaching leads to improvements in students’ knowledge/skills (i.e. test scores). We can use the results from our sample of students to draw conclusions about the impact of this module in general.

Let x = test score before the module, y = test score after the module

To test the null hypothesis that the true mean difference is zero, the procedure is as follows:

  1. Calculate the difference (di = yi − xi) between the two observations on each pair, making sure you distinguish between positive and negative differences.
  2. Calculate the mean difference, d¯.
  3. Calculate the standard deviation of the differences, sd, and use this to calculate the standard error of the mean difference, SE( d¯) = √sd n
  4. Calculate the t-statistic, which is given by T = d¯ SE( d¯) . Under the null hypothesis, this statistic follows a t-distribution with n − 1 degrees of freedom.
  5. Use tables of the t-distribution to compare your value for T to the tn− 1 distribution. This will give the p-value for the paired t-test.

NOTE:

For this test to be valid the differences only need to be approximately normally distributed. Therefore, it would not be advisable to use a paired t-test where there were any extreme outliers.

Example

Using the above example with n = 20 students, the following results were obtained:

Student Pre-module Post-module Difference score score 1 18 22 + 2 21 25 + 3 16 17 + 4 22 24 + 5 19 16 - 6 24 29 + 7 17 20 + 8 21 23 + 9 23 19 - 10 18 20 + 11 14 15 + 12 16 15 - 13 16 18 + 14 19 26 + 15 18 18 0 16 20 24 + 17 12 18 + 18 22 25 + 19 15 19 + 20 17 16 -

Calculating the mean and standard deviation of the differences gives:

d¯ = 2.05 and sd = 2.837. Therefore, SE( d¯) = √sd n

=^2 √.^837

So, we have:

t =

= 3. 231 on 19 df

Looking this up in tables gives p = 0.004. Therefore, there is strong evidence that, on average, the module does lead to improvements.