Question

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

Scenario

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	Comparison District 1	Comparison District 2
332	324	301
317	307	317
320	277	297
344	333	303
314	300	320
330	302	309
291	320	278
302	291	299
286	296	290
318	327	311
320	304	314
327	277	297
283	284	264
322	276	270

Questions

1. What is the mean scale score for each district?

Superintendent’s District=314.71

Comparison District 1=301.29

Comparison 2=297.86

2. What is the standard deviation of the scale scores for each district?
3. State an appropriate null hypothesis for this analysis.
4. What is the observed or computed value of F?
5. What are the reported degrees of freedom for the Between Groups variance?
6. What are the reported degrees of freedom for the Within Groups variance?
7. What is the reported level of significance?
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.
10. Present the results as they might appear in an article. This must include a table and narrative statement that reports the results of the one-way ANOVA and LSD post hoc test.

Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.

Answer 1

1.standard deviations od scale scores for each district

standard deviation of superintendent district

=4298.02

σ =

= 17.52

standard deviation of comparison district 1

=4826.80

=

=18.56

standard deviation of comparison district 2

= 3831.58

=

=16.54

2. Null hypothesis for this analysis

H0 : There is no difference in performance on high stakes tests amng the districts

alternative hypothesis ( H1) :  There is difference in performance on high stakes tests amng the districts.

Source of variation	Sum of squares	Degrees of freedom	Mean sum of squares	F statistic
Between sum squares Within sum of squares	2220.32 12956.4	3-1=2 42-3=39	2220.32/2 =1110.16 12956.4/39 =332.21	1110.16/332.21 =3.32

within sum of squares =

where i = 1,2, 3

3. the observed or computed value of F

Fcal = 3.34

4.the reported degrees of freedom for the Between Groups variance

K-1

where k is number of conditions in the our case ( number of districts ) =3

3-1 =2

degrees of freedom for the Between Groups variance is 2

5. .the reported degrees of freedom for the within Groups variance

N-K = 42-3 = 39

where N is total number of obersevations

6 . level of significan is 0.5