## Question 526:

1

1. Q526_1_2tQ526.xls
1. The Null hypothesis is the grades turned in earlier are not better than grades from later tests.  The Alternative hypothesis is grades are higher for tests turned in earlier.
2. 2. For this test we'd use a 2-sample t-test. The test-statistic is generated by subtracting the means and dividing by the standard deviation of the sample. We are asked to assume equal variances so we need to generate the estimate of this standard deviation using the following procedure.
3. Since the sample sizes are unequal, as well as the standard deviations, we generate a pooled standard deviation using the variance (squaring the standard deviation) and the sample size.
1. In assuming unequal variance we use the formula: [ (n1 -1)var1 + (n2 -1)var2  ] /  ( n1 +  n2 -2 )
1.  =[ (25 -1)384.16 + (24 -1)620.01  ] / ( 25 +  24 -2 )
2.  = SQRT(499.576)
3. = 22.3512 which is the pooled standard deviation
4. The standard deviation of the mean difference (also called the standard error of the mean difference) is calculated my multiplying the standard deviation found above times the square root of (1/ n1 + 1/ n2)  = 22.3512* SQRT(1/25+/1/24) = 6.387
2. The difference between the means is 77.1-69.3= 7.8 making the test statistic = 7.8/6.387= 1.221
3. There  are 47 degrees of freedom (n1+n2-2).
4. Looking up the p-value associated with this t-statistics and 47 degrees of freedom gets us a p-value of .114.  We can also use the Excel function =TDIST(1.221,47,1), where the parameters are the t-statistics, degrees of freedom and 1 tailed test.
4. The p-value of .114 is above our rejection criteria of .05 meaning we would fail to reject the null hypothesis. This means we would not be able to conclude the difference observed is greater than chance, meaning there is no difference in grades when they are turned in earlier.
5. The observed difference in means test scores between the early tests and later tests was 7.8 or about (77.1-69.3)/69.3 = 11.8% better for the first group. To me, an almost 12% difference can mean the difference between passing or failing or between a B and C, so I think that difference is meaningful.
6. We assumed equal variances. To test this using just the standard deviations we can use the F-test, which is a ratio of each variance.
1. We calculated each variance above and their ratio 620.01/384.16 =  1.613.
2. The null hypothesis is that there is no difference between the variances.
3. Using the excel function =FDIST(1.6139,23,24), where the parameters are the ratio (F) and 1 less than each sample size. This provides us with a p-value of .1256, which is the 1-sided area. We multiply this value times 2 and get our p-value of .251.
4. Since the p-value of .251 is above the rejection criteria of .05, we would NOT reject the null hypothesis and conclude the variances are not different.

A final note on testing variances. There are multiple ways to test for unequal variances, the F-test is one way however it has been criticized as being too sensitive to outliers in the data (since the standard deviation is based on the mean and both are therefore affected by outliers). Other tests, such as the Levine test are recommended, however the raw data are needed for this test.

See the attached Spreadsheet for the calculations