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,

Chart Title 60 0 2 3 et LL

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,

Chart Title 60 50 40 30 20 10 0 2 ПА ПВ шс D ВЕ et ПС в а

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.

2 g= L4

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,

A C E Maze variety 1 2 1 2 3 4. 46 3 A 31 35 38 4 B 29 32 45 36 5 C 13 17 35 20 6 D 7 E 28 38 52 39 14 20 40 20 LL CC

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

ME PAGE LAYOUT INSERT FORMULAS DATA REVIEW Data Analysis ZA A Z Data Analysis X Analysis Tools OK e Anova: Single Factor Anova: Two-Factor With Replication Anova: Two-Factor Without Replication Correlation Cancel Help Covariance Descriptive Statistics Exponential Smoothing F-Test Two-Sample for Variances Fourier Analysis Histogram 29 22 45 26. L4

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

A B C E F Maze variety.- 1 41 2 1 2 33 3 A 38i 31 35 46 4 B 29 32 45 36! 20 39 5 13 17 35 6 D 28 38 52 7 E 201 14 20 40 8 Anova: Two-Factor Without Replication ? X 10 Input OK SAS2:SES7 11 Input Range: Cancel 12 Labels 13 Alpha: 0.05 Help 14 Output options 15 O Output Range: SOS1 16 New Worksheet Ply: 17 18 New Workbook 19 Ln C LO LO

The result is obtained.  The screenshot is shown below,

H K M 16 ANOVA F crit 17 Source of Variation df P-value SS MS 18 Rows 1124.3 4. 281.075 44.79283 4.05E-07 3.259167 19 Columns 1145.2 3 381.7333 60.834 1.57E-07 3.490295 20 Error 75.3 12 6.275 21 22 Total 2344.8 19

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.