Question

Conduct an analysis (you can use Excel) to answer the following question. Show an ANOVA table.

A grower wishes to compare yields of four different varieties of pumpkins for yield (tons/ac).  His experiment consisted of four replications for each variety.  Is there a significant difference between the varieties in yield?  Do a Tukey’s test. Give complete conclusions with a properly labeled tabulated results. Data are as follows:

Variety A        Variety B        Variety C       Variety D

27                   29                   30                   32

28                   28                   29                   29

29                   29                   29                   31

25                   27                   31                   32

An experiment was conducted to compare the effect of four different irrigation treatments (1, 2, 3, 4) on grape yield (lbs/vine).  Results obtained were as follows:

Treatment 1	47	52	62	51
Treatment 2	50	54	67	57
Treatment 3	57	53	69	57
Treatment 4	54	65	75	59

Conduct an ANOVA in Excel to verify if the treatments resulted in significant differences in grape yield.  Conduct an LSD test to find out which treatments were the best for producing higher grape yields, if the ANOVA was significant. Make appropriate conclusions and present your results in a graph.

3.  A sample of overwintering lady-bird beetles consisted of 37 “black with red spots” phenotypes and 42 “red with black spots” phenotypes.  Test the hypothesis that the overwintering populations consists of equal numbers of each phenotype. Do the calculation by hand

Answer 1

3)

From excel you get the following output for one-way ANOVA -

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Variety A	4	109	27.25	2.916666667
Variety B	4	113	28.25	0.916666667
Variety C	4	119	29.75	0.916666667
Variety D	4	124	31	2
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	32.6875	3	10.89583333	6.456790123	0.007528174	3.490294819
Within Groups	20.25	12	1.6875
Total	52.9375	15

As the p-value of test is less than 0.05, so we reject the null hypothesis and conclude that there is enough evidence in the data to support the claim that there is significant difference in at least one pair of means.

Tukey's test -

The critical value of Q-statistic for degree of freedom of 12 at significance level of 0.05 is -

Q_{4, 12, 0.05} = 4.2

So, critical value for difference in mean is -

So, if the difference between any two means has magnitude greater than 2.7280, we would say that the difference is significant otherwise not significant.

We can create following table -

	Tukey's HSD
	Mean Difference	Critical Difference	Significant
Variety A - Variety B	1	2.728	No
Variety A - Variety C	2.5	2.728	No
Variety A - Variety D	3.75	2.728	Yes
Variety B - Variety C	1.5	2.728	No
Variety B - Variety D	2.75	2.728	Yes
Variety C - Variety D	1.25	2.728	No

So, we conclude that the yield of variety D is significantly different from the yield of both the varieties A and B.

And as the mean of D is higher than both A and B, so we can say that the mean yield of D is significantly higher than mean yield of both A and B while all the other yields are not significantly different.

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Please ask other questions separately.