Module 10: Hypothesis Testing With Two Samples
Matched or Paired Samples
Barbara Illowsky & OpenStax et al.
When using a hypothesis test for matched or paired samples, the following characteristics should be present:
- Simple random sampling is used.
- Sample sizes are often small.
- Two measurements (samples) are drawn from the same pair of individuals or objects.
- Differences are calculated from the matched or paired samples.
- The differences form the sample that is used for the hypothesis test.
- 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:
[latex]displaystyle{t}=frac{{overline{{x}}_{{d}}-{mu}_{{d}}}}{{{(frac{{s}_{{d}}}{sqrt{{n}}})}}}[/latex]https://www.youtube.com/embed/5ABpqVSx33I
Example
A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the table below. 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 |
Solution:
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 [latex]displaystyleoverline{{x}}_{{d}}=-{3.13}{quadtext{and}quad}{s}_{{d}}={2.91}[/latex]. Verify these values.
Let be the population mean for the differences. We use the subscript to denote “differences.”
Random variable: [latex]displaystyleoverline{{X}}_{{d}}[/latex] = 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 hypnotism. 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 hypnotism. That means the subject shows im