1. Introduction

This work primarily concerns itself with the application of test hypothesis with directional alternatives. This has application in many fields. For examples, there may be a number of cures for a particular ailment. Orthodox chemotherapy, non-orthodox herbal and “body cure” (allowing the body to perform the cure) may be three methods of cure. The best, in terms of duration of treatment before cure could be chemotherapy, followed by the herbal. Thus, if time to realize the cure is the concerned variable, then the means would be ordered from the shortest (obtained from orthodox) to the one obtained by herbs, while the time for body cure is the last. Several authors have provided adequate literature in this area of Analysis of Variance such as Ott (1984), Montgomery (1981) amongst. Many authors have also developed method of testing of homogeneity of means. Such authors include Jonckheere (1954), Bartholomew (1959), Yahya and Jolayemi (2003), Dunnett (1964), Barlow et. al (1971), Dunnett (1980), Dunnett and Tamhane (1997), Keselman and Wilox (1999), Gupta et al (2006), Abidoye et. al (2013a), (2013b), (2013c) and several others.

2. Development of the Test Procedure

In this section we are interested in developing a suitable test procedure to test the hypothesis against ordered means which is given by

(2.1)

[The situation in which the hypothesis set is given as

vs can also be handed similarly]

When then the alternative hypothesis is sustained. In this formulation then,

(2.2)

Now assume that and identically and independently distributed where for at least one i, i =1, 2, …, g, j = 1, 2, …, n_i

The unbiased estimate of

(2.3)

and that

(2.4)

so that

(2.5)

where

(2.6)

obtained from order statistics, from equation (2.5), will be appropriate to be used as the estimate if the group of variances are equal, i.e where

Consider the observations in which there exists at least one outlier. Within this situation it is a well known fact that sample mean given as

is not a good location summary. Instead, the Harmonic mean given by

is a better choice. Similarly let be the variances of the g groups which is the Harmonic mean of the variances is a better choice whenever at least one .

Note that

see abidoye et. al (2012, 2013a and 2013b)

But in a situation where the group variances are not equal, that is, for at least one (pooled variance which is the weighted mean of variances) can not be used. This is the Behren’s – Fisher problem.

Thus

The test statistic for the hypotheses set in (2.2) is therefore

(2.7)

where

(2.8)

and

(2.9)

where is the appropriate normalization from order statistics. The null hypothesis is then rejected if

(2.10)

where t_r is the regular t – distribution with r d.f is given by

because has approximately the - distribution with the degree of freedom r to be determined from r = 22.096 + 0.266(n-g) - 0.000029 . See Abidoye (2012) and Abidoye et. al 2013a. If the alternative hypothesis (H₁) in equation (2.10) is reversed.

Then

(2.12)

3. Application of the Test Procedure

Practical situations where this test procedure is applicable are presented in this section. The data on Maize was found in Kwara Agricultural Development Project (KWADP), Ilorin, Kwara State for the period of 1998 to 2007. By KWADP standard, Kwara State has been categorized into four major unique ecological zones based on the climatic and ecological differences.

3.1. Example on Maize Yields in Kwara State

Consider the data on yields of maize (in Tonnes per hectare) given by KWADP who believed that crop yield in the area under consideration, zone D has the highest yield followed by zone A, then zone C and zone B with the least yield.

If represents the mean population yield of maize in zone i where

Let represents the crop (maize) yield (in Tonnes per hectare) in zone i for year j, i = 1, 2, 3, 4 ; j=1, 2, …, 10.

Then, assume

(2.13)

First it is necessary to examine the values of the group variances. This was done in Table 3.1, confirming unexqual variances.

Table 3.1. Spss Result Output on Test of Homogeneity of Variances For Maize

The hypothesis to be tested is

Vs ; for at least one i

The Proposed Test Statistic is

where

and

The null hypothesis, H₀ is rejected at specified α level of significance if

P-value = , where

COMPUTATION ON MAIZE

From the data on Maize yield the following summary statistics were obtained:

Zone A: =1.014, =0.056, =10

Zone B: =0.618, =0.0126, =10

Zone C: =0.8, =0.0111, =10

Zone D: =2.776, =0.027, =10

Therefore,

Then consider the differences given below:

= 2.776 - 1.014 = 1.762

= 1.014 - 0.80 = 0.214

= 0.80 - 0.618 = 0.182

with , g = 4

where r = 22.096 + 0.266(n-g) - 0.000029

= 31.634

, 0 < λ< 1 from equation (2.1)

Which led to the rejection of H₀ and conclude that the mean yields of maize in all the four zones were indeed ordered from D, A, C, B._.

4. Conclusions

In this study, we examined a test statistic for testing ordered alternative hypothesis under unequal variances. We proposed a modified t – test statistic, the harmonic mean of the variances to replace the sample pooled variance so as to avoid the Beheren – Fisher’s problem. Because the sample harmonic mean of variances has the chi – square distribution, then the modified t – statistic is found to be appropriate for the data set used from Kwara Agricultural development project (KWADAP).