Matched or Paired Samples Quiz

PDF page 593: 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: ̄ x − μ

PDF page 593: A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.10. 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.

PDF page 593: 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 = –3.13 and s = 2.91 Verify these d d

PDF page 593: 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 = –3.13 and s = 2.91 Verify these d d

PDF page 594: 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 improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

PDF page 594: Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t7. (Notice that the

PDF page 594: test is for a single population mean.) Calculate the p-value using the Student's-t distribution: p-value = 0.0095 Graph:

PDF page 594: 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. NOTE 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

PDF page 595: 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:

PDF page 595: 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. 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. ̄

PDF page 596: ̄ Define the random variable: X mean difference in the maximum lift per player. d The distribution for the hypothesis test is t3.

PDF page 596: H0: μd ≤ 0, Ha: μd > 0