3. Two teachers are up for a promotion, but their district will only give it to one of them if there is a significant difference in their teaching effectiveness. They decide to determine if one of them is a better teacher by having them give their students a common final. The results are summarized in the following table.

 number of students mean score standard deviation Teacher A 27 79 5.2 Teacher B 35 84 6.3

Is there a significant difference between the two teachers?

• a) State the null hypothesis and the alternative hypothesis
• b) Is there a significant difference at the .05 level of significance?

a) The null hypothesis is that there is no difference between the two teachers.

The alternative hypothesis is that there is a difference between them. Since we are trying to determine which one is better, we have no preconceived ideas as to which way the test should go, so this is a two tailed test.

b) To tell if there is a significant difference we will use a two tailed t-test. The t statistic is computed using the following formula

We substitute our numbers and get

In order to compare this to our critical value of t, we need to find the number of degrees of freedom. We use the foumula,

Substituting in our numbers we get

= 59.69

The closest number of degrees of freedom in Table D would be 60 degrees of freedom. Since this is a two tailed test at a 5% level of significance, there will be 2.5% in each tail of the bell shaped curve, so we look under .025 above 95% at 60 degrees of freedom in Table D and we get a critical value

t* = 2.00

The t score which we computed of -3.42 has a larger absolute value than the critical valule of t, so we conclude that there is a significant difference between the two teachers. Teacher B is significantly better than teacher A at a 5% level of significance, so Teacher B should get the promotion according to this method of determination and Teacher A would not get promoted.