Question

A farmer has four different varieties of maize, and wants to compare their efficacy. He
recognises however that output produced depends not only on the variety of maize, but also
on the plot of land used for growing. He arranges an experiment in which quantity of each
variety is planted on each of the five one acre plots. This means that the sample size n is 5
while the number of treatments K is 4.
The yield is counted and shown in the table below.
MAIZE VARIETY (Treatment)

PLOT	1	2	3	4
A	31	35	46	38
B	29	32	45	36
C	13	17	35	20
D	28	38	52	39
E	14	20	40	20

a) Present the above information in diagram form using two appropriate methods.
10 marks
b) Formulate and test the hypotheses to determine whether there are any differences
between the varieties of maize, at 0.01 level of significance (hint: this is a two-way
ANOVA). 15 marks

Answer 1

a)

The data values are presented by two chart Line chart and the clustered column,.The charts are obtained in excel.

The first chart is line chart. The screenshot for the line chart is shown below,

From the chart we can see that difference in means is larger for group 2 and 3.

The second chart is the clustered column chart The screenshot for the line chart is shown below,

The chart also shows that the difference in means is larger for group 2 and 3.

b)

For varieties of maize

The null and alternative hypothesis are defined as,

H0: All the means are equal for varieties of maize, i.e.

H1: Atleast one of means is different.

A two factor ANOVA is used in excel to test the null hypothesis that all the means are equal for varieties of maze. The test is performed in excel by using following steps,

Step 1: Write the data values in excel. The screenshot is shown below,

Step 2: DATA > Data Analysis > ANOVA: Single Factor > OK.  The screenshot is shown below,

Step 3: Select Input Range: All the data values column. The screenshot is shown below,

The result is obtained.  The screenshot is shown below,

From the ANOVA table,

Maze varieties is described by the column

Source of Variation	P-value
Columns	1.57E-07	<	0.01	Reject the null hypothesis

Since the p-value is less than 0.01 at 5% significant level. The null hypothesis is rejected. Hence we can conclude that there is significant difference among varieties of maze.