Dependent samples t-test

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:

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.

Solution:

Corresponding “before” and “after” values form matched pairs. (Calculate “after” – “before.”)

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 3.13 and 2.91. Verify these values.

Null hypothesis:

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.)

Alternative hypothesis:

Ha: μd < 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 t7. (Notice that the test is for a single population mean.)

Calculate the p-value using the Student’s-t distribution: p-value = 0.0095

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 H0. This means that μd < 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.

EXAMPLE 2

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 this table.

Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.

Solution:

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 (3.71) and the sample standard deviation (4.5).

Calculate the p-value: The p-value is 0.0716 (using the data directly).

(test statistic = 2.18. p-value = 0.0719 using

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.


Concept 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

  • 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):

References:

  1. https://courses.lumenlearning.com/introstats1/chapter/matched-or-paired-samples/

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