Matched or Paired Samples (Optional) Quiz

PDF page 561: 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

PDF page 561: 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

PDF page 561: 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

PDF page 561: 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

PDF page 561: Problem A study was conducted to investigate the effectiveness of pain-reducing medication. Results for randomly selected subjects are shown in Table 10.11. 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 the medication? Test at a 5 percent significance level.

PDF page 561: Solution Corresponding before and after values form matched pairs. (Calculate after – before.)

PDF page 562: is the random variable for the differences. The sample mean and sample standard deviation of the differences are as follows: = –3.13 = 2.91

PDF page 562: is the random variable for the differences. The sample mean and sample standard deviation of the differences are as follows: = –3.13 = 2.91

PDF page 562: The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. 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 taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, 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. Note —that the test is for a single population mean. Calculate the p-value using the Student’s-t distribution: p-value = 0.0095 Graph:

PDF page 562: The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. 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 taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, 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. Note —that the test is for a single population mean. Calculate the p-value using the Student’s-t distribution: p-value = 0.0095 Graph:

PDF page 562: Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value.

PDF page 562: Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after taking the medication. The medication appears to be effective in reducing pain.

PDF page 562: Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after taking the medication. The medication appears to be effective in reducing pain.