Question

TABLE 1

An agronomist wants to compare the crop yield of 3 varieties of chickpea seeds. She plants 15 fields, 5 with each variety. She then measures the crop yield in bushels per acre. Treating this as a completely randomized design, the results are presented in the table that follows.

Trial	Smith	Walsh	Trevor
1	11.1	19.0	14.6
2	13.5	18.0	15.7
3	15.3	19.8	16.8
4	14.6	19.6	16.7
5	9.8	16.6	15.2

A. Referring to table 1, the agronomist decided to perform an ANOVA F test. The amount of total variation or SST is _____.

1. 82.39

2. 114.82

3. 32.43

4. 41.19

B. Referring to Table 1, the value of MSA is ____.

1. 114.82

2. 41.19

3. 82.39

4. 2.70

C. Referring to Table 1, the null hypothesis will be rejected at a level of significance of 0.01 if the value of the test statistic is greater than ___.

1. 15.24

2. 5.99

3. 6.93

4. 0.000508

D. Referring to Table 1, the decision made at .005 level of significance implies that all 3 means are significantly different.

1. True

2. False

3. Cannot Decide

Answer 1

For the given data Anova Single factor in Excel we get output as below

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Smith	5	64.3	12.86	5.463
Walsh	5	93	18.6	1.74
Trevor	5	79	15.8	0.905
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	82.38533333	2	41.19267	15.24149	0.000508	6.926608
Within Groups	32.432	12	2.702667
Total	114.8173333	14

So om ho above SST S HSA S Lu 19 t1L82 义