Question

(50%) 2, Four new brands (A. B. C. and D) of fertilizer that a farmer can use to grow crops just came on the market. Before deciding which brand he should use permanently for all crops, a farmer decided to experiment for one season. To do so, he randomly assign each fertilizer to five one acre tracts of land that he used to grow wheat. The following table gives the production of wheat (in bushels) for each acre for four brands of fertilizer Fertilizer A Fertilizer B Fertilizer C Fertilizer D 73 70 71 80 84 71 76 71 16.5 81 Sample mean Sample variance Population meanA 83 13.5 16.0 15.0 (10%) (a) Complete the following ANOVA table. At the 5% significance level, can you conclude that the mean yields of wheat are the same for these four brands of fertilizer? SOURCE BETWEEN BRANDS ERROR TOTAL DF MS

Answer 1

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D)

In this case we want to perform pairwise comparison of means. That is we want to test hypothesis of the sort:

$\dpi{100} H_0: \mu_A = \mu_B, H_0: \mu_A = \mu_C, H_0: \mu_A = \mu_D$ etc.

To perform the LSD method for the first hypothesis

$\dpi{100} H_0: \mu_A = \mu_B$ we will calculate $\dpi{100} LSD_{AB} = t_{0.05/2,DFW}\sqrt{MSW(1/n_A + 1/n_B)}$

where DFW = DF within and MSW = MS within
These values can be determine by performing ANOVA using the given information:

n	5	5	5	5
	A	B	C	D
1	73	68	75	80
2	70	77	82	87
3	71	76	82	86
4	65	73	80	84
5	76	71	86	78

					a
ANOVA Table					5%
Source	SS	df	MS	F	Fcritical	p-value
Between	520	3	173.33	11.366	3.2389	0.0003	Reject
Within	244	16	15.25
Total	764	19

From the ANOVA table, the MS-within = 15.25

DF-within = 16

Therefore $\dpi{100} LSD_{AB} = t_{0.05/2,DFW}\sqrt{MSW(1/n_A + 1/n_B)}$

= $\dpi{100} LSD_{AB} = t_{0.05/2,16}\sqrt{15.25(1/5 + 1/5)}$

= 2.120*sqrt(6.1) = 5.2360

In this case since n_A = n_B = n_C = n_D = 5, therefore $\dpi{100} LSD_{AB} = LSD_{AC} = ... = LSD_{CD}$ and we compare all pairwise differences in the mean to 5.2360

1. Pair 1: If $\dpi{100} |\bar{y}_A-\bar{y}_B| \geq LSD_{A,B}$ , then we reject the null hypothesis $\dpi{100} H_0: \mu_A = \mu_B$ . In this case

$\dpi{100} |\bar{y}_A-\bar{y}_B| = 2$ is less than LSD (5.2360) therefore we do not reject the null hypothesis at the 0.05 level. Similarly we will compare all pair mean differences:

$\dpi{100} |\bar{y}_A-\bar{y}_B| = 2 < 5.2360$ , do not reject $\dpi{100} H_0: \mu_A = \mu_B$

$\dpi{100} |\bar{y}_A-\bar{y}_C| = 10 > 5.2360$, reject $\dpi{100} H_0: \mu_A = \mu_C$

$\dpi{100} |\bar{y}_A-\bar{y}_D| = 12 > 5.2360$, reject $\dpi{100} H_0: \mu_A = \mu_D$

$\dpi{100} |\bar{y}_B-\bar{y}_C| = 8 > 5.2360$, reject $\dpi{100} H_0: \mu_B = \mu_C$

$\dpi{100} |\bar{y}_B-\bar{y}_D| = 10 > 5.2360$, reject $\dpi{100} H_0: \mu_B = \mu_D$

$\dpi{100} |\bar{y}_C-\bar{y}_D| = 2 < 5.2360$, do not reject $\dpi{100} H_0: \mu_C = \mu_D$

NOTE:According to my knowledge i have answered only c,d and f bits. if u want remaining bits please post to another window.if you see my answer please rate .THANKS