Question

1. The following data is on seedling growth for tomato plants. The data are on the increase in height after 16 days of planting with different numbers of nematodes in each pot. Is there a significant difference in growth at the 5% significance level?

Nematodes Seedling growth

0 10.8 9.1 13.5 9.2

1,000 11.1 11.1 8.2 11.3

5,000 5.4 4.6 7.4 5.0

10,000 5.8 5.3 3.2 7.5

that is all the problem given shows have to decide if i to use anova or chi squared

Answer 1

Since you need to find if there is a significant difference among the 4 groups, you need to perform a one way ANOVA.

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The sample statistics from the above data is as below:

	0'	1000'	5000'	10000'
Total	42.6	41.7	22.4	21.8
n	4	4	4	4
Mean	10.650	10.425	5.6	5.45
SS	12.65	6.6275	4.64	9.41

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The Hypothesis:

H0: There is no difference between the mean of the four treatments.

Ha: The mean of at least one treatment is different from the others.

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The Test Statistic:

.The ANOVA table is as below.

The critical value is calculated for = 0.05 for df1 = 3 and df2 = 12; Critical value = 3.4903

The p value is calculated for F = 40.127 for df1 = 3 and df2 = 12; p value = 0.0006

Source	SS	DF	Mean Square	F	Fcv	p
Between	100.65	3.00	33.55	12.08	3.4903	0.0006
Within/Error	33.33	12.00	2.78
Total	133.98	15.00

F_observed = 12.08

The Decision Rule: If F_observed is > F_critical, Then reject H0.

Also if p-value is < , Then reject H0.

The Decision: Since Fobserved (12.08) is > F_critical (3.4903), We reject H0.

Also since p-value (0.0006) is < (0.05), We reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that the mean of at least one treatment is different from the others.

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Calculations For the ANOVA Table:

Overall Mean = [42.6 + 41.7 + 22.4 + 21.8] / 4 = 8.0313

SS treatment = SUM n* ( - overall mean)² = 4 * (10.65 - 8.0313)² + 4 * (10.425 - 8.0313)² + 4 * (5.6 - 28.0313)² + 4 * (5.45 - 8.0313)² = 100.65

df1 = k - 1 = 4 - 1 = 3

MSTR = SS treatment / df1 = 100.65 / 3 = 33.55

SSerror = SUM [Sum of Squares (SS)] = 12.65 + 6.6275 + 4.64 + 9.41 = 33.33

df2 = N - k = 16 - 4 = 12

Therefore MS error = SSerror / df2 = 33.33 / 12 = 2.78

F = MSTR/MSE = 33.55 / 2.78 = 12.08

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