Question

Assignment #5: Analysis of Variance and Post Hoc Tests
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:
298 301 326 313 314 310 310 313 320 304 304 301 299 310

Comparison district 1:
290 288 289 294 307 310 320 274 286 301 280 295 296 310   

Comparison district 2:

302 277 298 306 287 292 309 306 300 285 309 278 290 295   


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

Groups	Average
Superintendent’s district	308.7857
district 1	295.7143
district 2	295.2857

2.

Groups	SD
Superintendent’s district	8.172925
district 1	12.73051
district 2	10.83746

3.

Null Hypothesis:

The null hypothesis states that all the mean scores are equal.
4.

Source of Variation	F
Between Groups	7.141141

5.

Source of Variation	df
Between Groups	2

6.

Source of Variation	df
Within Groups	39

7.

Source of Variation	P-value
Between Groups	0.002277

8.

P-value < 0.05 at a 5% significance level hence the null hypothesis is rejected.

9.

Superintendent’s district is significantly different from district 1 and district 2.

Explanation:

A single factor ANOVA is used in excel to test the null hypothesis that all the means are equal. The hypothesis is defined as,

Null Hypothesis:

Alternative hypothesis; Alt least one mean is significantly different.

The test is performed in excel by using the following steps,

Step 1: Write the data values in excel. The screenshot is shown below,

Step 2: DATA > Data Analysis > ANOVA: Single Factor > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column, Grouped By: Columns. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

From the result summary,

P-value Approach: P-value < 0.05 at a 5% significance level hence the null hypothesis is rejected. Hence, we can state that there is a statistically significant difference among means at a 5% significance level.

Post Hoc Test

Since there are differences among the groups, Fisher's LSD approach is used to identify the groups that significantly different in the mean.

The LSD value is obtained using the formula,

Where,

(Obtained from ANOVA result summary,)

Decision Rule: Reject the null hypothesis if difference in means are greater than LSD value,

Now compare the difference in means with the LSD,

Comparison	Difference in means	LSD
Superintendent’s district vs district 1	13.07143	>	8.2140	Significant
Superintendent’s district vs district 2	13.5	>	8.2140	Significant
district 1 vs district 2	0.428571	<	8.2140	Not significant