Question

Directions: Using the information from the following scenario, conduct a one-way ANOVA and specify the LSD post hoc test.

The superintendent is continuing to examine the data that has been reported for the district. Another question concerned the differences in performance on high stakes tests. To examine this issue, the superintendent obtained the average scale scores for schools that participated in the high stakes testing for the district and two comparison districts. The following scores were collected:

Superintendent’s district:

307    318    325    313    312    334    335    331    336    304    336    328       

Comparison district 1:

299    345    323    286    314    286    283    270    308    272    291    297     

Comparison district 2:

287    299    332    300    311    299    317    309    307    304    299    313

8. Based on the results of the one-way ANOVA, would you accept or reject the null hypothesis?

9. Based on the results of the LSD post hoc test, identify which districts are significantly different.

Answer 1

The null and alternative hypothesis for one way ANOVA are:

H₀: There is no difference in performance on high stakes test.

H₁: There is a difference in the performance that is at least one means differ in the group.

F test statistics

To find MST and MSE, first, find the sum of square for treatment that is SST and sum of squares for error that is SSE.

j is for column there are 3 columns.

and n1 = n2 = n3 = 12

and

That is subtract column mean from each column observations, there are 12 observation for the first column that is for Superintendent's district and same for the remaining two columns and then the square of them and then add all 36 squared to get SST.

There are 3 treatment groups so DF for treatment = k - 1 = 3 - 1 = 2

Anf there are total 36 observations, so DF for error = n - k = 36 - 3 = 33

To find F critical value,

Level of significance is not given so take it as 0.05, DF of numerator = 2 and DF for denominator = 33

By using the F distribution table the F critical value is 3.285

Decision rule:

If F test statistics < F critical value then we fail to reject the null hypothesis otherwise reject the null hypothesis.

Here F test statistics (8.074) is not less than F critical (3.285) So we reject the null hypothesis.

Conclusion: There is a difference in the performance.

LSD post hoc test:

Where t is the t critical value at DF of MSE that is DF = 33

By using t distribution table, the t critical value for two tail with DF = 33 is 2.0345

MSE = 248.4495 and n = 12 that is number of observation per cell

Now find the absolute differences of means that is first and second, first and third and second and third group.

The absolute mean difference between the first and second group = 25.417

The absolute mean difference between the first and third group = 16.8333

The absolute mean difference between the second and third group = 8.5837

Now compare LSD with all mean difference, the first two are greater than LSD but the last group(8.5837) is less than the LSD.

So, the second and third group differs from the rest.

That is there is a significant difference between the Comparision district 1 and comparison district 2.