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 [latex]\displaystyle\overline{{x}}_{{d}}=-{3.13}{\quad\text{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]\displaystyle\overline{{X}}_{{d}}[/latex] = the mean difference of the sensory measurements

H₀: μ_d ≥ 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. μ_d is the population mean of the differences.)

H_a: μ_d < 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 (In Excel, use =T.TEST(array1, array2, 1, 1)=0.0095)

Graph:

[latex]\displaystyle\overline{{X}}_{{d}}[/latex] is the random variable for the differences.

The sample mean and sample standard deviation of the differences are:

[latex]\displaystyle\overline{{x}}_{{d}}=-{3.13}[/latex]

[latex]\displaystyle{s}_{{d}}={2.91}[/latex]

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 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 after hypnotism. Hypnotism appears to be effective in reducing pain.

Example

A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a 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:

The coach wants to know if the strength development class makes his players stronger, on average.

Solution:

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: {90, 11, -8, -8}. Assume the differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation.

[latex]\displaystyle\overline{{x}}_{{d}}={21.3},{s}_{{d}}={46.7}[/latex]

Note: 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 mean.

Define the random variable: [latex]\displaystyle\overline{{X}}_{{d}}[/latex] mean difference in the maximum lift per player.

The distribution for the hypothesis test is t₃  (there are 4 – 1 or 3 degrees of freedom).

H₀: μ_d ≤ 0, H_a: μ_d > 0

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.

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.

Example

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 and the sample standard deviation. [latex]\displaystyle\overline{{x}}_{{d}}={3.71},{s}_{{d}}={4.5}[/latex].

Random variable: [latex]\displaystyle\overline{{X}}_{{d}}[/latex] = mean difference in the distances between the hands.

Distribution for the hypothesis test:t₆

H₀: μ_d = 0 H_a: μ_d ≠ 0

Graph:

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

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.