Matched or Paired Samples Quiz

PDF page 444: differences exist. Each observation has its matched pair against which differences are calculated. First, the differences in the metric to be tested between the two lists of observations must be calculated, and this is typically labeled with the letter ̄ "d." Then, the average of these matched differences, X is calculated as is its standard deviation, Sd. We expect that the d standard deviation of the differences of the matched pairs will be smaller than unmatched pairs because presumably fewer differences should exist because of the correlation between the two groups. When using a hypothesis test for matched or paired samples, the following characteristics may be present:

PDF page 449: A hypothesis test for matched or paired samples (t-test) has these characteristics:

PDF page 444: 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, that is, the number of pairs not the number of observations. The null and alternative hypotheses for this test are:

PDF page 449: • Distribution: Student’s-t distribution with n – 1 degrees of freedom

PDF page 444: 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, that is, the number of pairs not the number of observations. The null and alternative hypotheses for this test are:

PDF page 445: n − 1 We can now compare the calculated value of the test statistic, 4.12, with the critical value. The critical value is a Student's t with degrees of freedom equal to the number of pairs, not observations, minus 1. In this case 20 pairs

PDF page 444: differences exist. Each observation has its matched pair against which differences are calculated. First, the differences in the metric to be tested between the two lists of observations must be calculated, and this is typically labeled with the letter ̄ "d." Then, the average of these matched differences, X is calculated as is its standard deviation, Sd. We expect that the d standard deviation of the differences of the matched pairs will be smaller than unmatched pairs because presumably fewer differences should exist because of the correlation between the two groups. When using a hypothesis test for matched or paired samples, the following characteristics may be present:

PDF page 444: A company has developed a training program for its entering employees because they have become concerned with the results of the six-month employee review. They hope that the training program can result in better sixmonth reviews. Each trainee constitutes a “pair”, the entering score the employee received when first entering the firm and the score given at the six-month review. The difference in the two scores were calculated for each employee and the means for before and after the training program was calculated. The sample mean before the training program was 20.4 and the sample mean after the training program was 23.9. The standard deviation of the differences in the two scores across the 20 employees was 3.8 points. Test at the 10% significance level the null hypothesis that the two population means are equal against the alternative that the training program helps improve the employees’ scores.

PDF page 444: A company has developed a training program for its entering employees because they have become concerned with the results of the six-month employee review. They hope that the training program can result in better sixmonth reviews. Each trainee constitutes a “pair”, the entering score the employee received when first entering the firm and the score given at the six-month review. The difference in the two scores were calculated for each employee and the means for before and after the training program was calculated. The sample mean before the training program was 20.4 and the sample mean after the training program was 23.9. The standard deviation of the differences in the two scores across the 20 employees was 3.8 points. Test at the 10% significance level the null hypothesis that the two population means are equal against the alternative that the training program helps improve the employees’ scores.

PDF page 445: = = 4.12 Test Statistic: tc = S ⎛ 3.8 ⎞ d ⎝ ⎠ n 20 In order to solve this equation, the individual scores, pre-training course and post-training course need to be used to calculate the individual differences. These scores are then averaged and the average difference is calculated: ̄ ̄ ̄

PDF page 445: and at 90% confidence level ta/2 = ±1.729 at df = 20 - 1 = 19. The calculated test statistic is most certainly in the

PDF page 445: tail of the distribution and thus we cannot accept the null hypothesis that there is no difference from the training program. Evidence seems indicate that the training aids employees in gaining higher scores.

PDF page 445: Solution 10.10 Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.")

PDF page 446: 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 values. Let μ be the population mean for the differences. We use the subscript d to denote "differences." d

PDF page 446: – X is the random variable for the differences. d The sample mean and sample standard deviation of the differences are: – x = –3.13 d

PDF page 446: 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 values. Let μ be the population mean for the differences. We use the subscript d to denote "differences." d

PDF page 446: – s = 2.91 d Compare the critical value for alpha against the calculated test statistic. The conclusion from using the comparison of the calculated test statistic and the critical value will gives us the result. In this question the calculated test statistic is 3.06 and the critical value is 1.895. The test statistic is clearly in the tail and thus we cannot accept the null hypotheses that there is no difference between the two situations, hypnotized and not hypnotized. Make a decision: Cannot accept the null hypothesis, H0. This means that μd < 0 and there is a statistically significant improvement.