1. Suppose we are comparing five different fertilizer treatments for tomatoes: A

Question

1.   Suppose we are comparing five different fertilizer treatments for tomatoes: A, B, C, D, and E. Each of 30 potted tomato plants is randomly assigned to one of the treatments, and the total yield of tomatoes for each plant is recorded at the end of the growing season. The data are summarized here:

a.   Fertilizers A and B and C are three different brands of synthetic fertilizer, with similar nutrients and concentrations. Fertilizers D, and E are different forms of compost: compost tea and solid compost. What linear combination should you use to test that the average of the mean yields using synthetic fertilizer is equal to the average of the mean yields using compost? Give the coefficients for the linear combination, and state the null hypothesis that you would be testing.

b.   What is the standard error (square root of estimated variance) of the linear combination in part a.?

c.   What is the t-statistic for testing the hypothesis in part a.?

d.   Find the resulting p-value and state your conclusion.

Fertilizer A B C D E All Combined Sample Size 3 8 5 6 8 30 Sample Mean 23.53 23.92 27.68 26.34 28.21 26.13 Sample Variance 1.30 4.96 4.07 0.63 1.25 5.91

Explanation / Answer

Solution

Back-up Theory

Let X = yield of tomatoes with synthetic fertilizer (A, B & C) and …………………..(1)

      Y = yield of tomatoes with compost (D &E) …………………………………….(2)

We assume X ~ N(µ1, 12) and Y ~ N(µ1, 12).. ………………………………………(3)

Now, to work out the solution,

Part (a)

Since sample sizes are different, all sample statistics should be grouped by an appropriate linear combination with sample size as weights.

The linear combination for synthetic fertilizer would be: (3/16)A + (8/16)B + (5/16)C.

The linear combination for compost would be: (6/14)D + (8/14)E

So, the coefficients are: (3/16), (8/16), (5/16) for synthetic fertilizer and (6/14), (8/14) for compost. ANSWER1

The null hypothesis is: H0: µ1 = µ2 [vide (3) under Back-up Theory] ANSWER2

Part (b)

Standard error of linear combination for synthetic fertilizer

= square root of [{(n1 - 1)s12 + (n2 - 1)s22+ (n3 - 1)s32}(n1 + n2 + n3 - 3)], where n1,n2 and n3 are the sample sizes for A, B and C and s12 ,s22 , s32 are the respective sample variances.

= sq.rt(53.6/13) = 2.0305 ANSWER1

Similarly,

Standard error of linear combination for compost

= square root of [{(n4 - 1)s42 + (n5 - 1)s52 }(n4 + n5 - 2)], where n4 and n5 are the sample sizes for D and E and s42 ,s52 are the respective sample variances.

= sq.rt(11.2/12) = 0.9661 ANSWER2

Part (c)

t-Statistics for testing H0: µ1 = µ2 is: t = |(Xbar – Ybar)|/[s.{(1/n) + (1/m)}] , where

Xbar = combined mean of A, B and C = {(3/16)(23.53) + (8/16)(23.92) + (5/16)(27.68)} = 25.02

Ybar = combined mean of D and E = {(6/14)(26.34) + (8/14)(28.21)} = 27.41

s = square root of [{(n1 - 1)s12 + (n2 - 1)s22+ (n3 - 1)s32 + (n4 - 1)s42 + (n5 - 1)s52}/ (n1 + n2 + n3 + n4 + n5 - 5)] = 1.61

n = n1 + n2 + n3 = 16 and m = 14.

Thus, t = - 2.39/0.5890 = - 4.058 and |t| = 4. 058 ANSWER

Part (d)

To perform the test

Calculated value of |t| = 4.058 [as obtained above]

The distribution of the statistic is t with 25 degrees of freedom.

The p-value of 4.058 = 0.0004 [using Excel Function], which is much less than the usual level of significance, 0.05. Hence the null hypothesis is rejected.

=> there is strong evidence to suggest that the two means are not equal. ANSWER

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