Case 9 Uniform Uniforms

Introduction

This case is concerned with the topic Analysis of variance (ANOVA). It is a statistical method of

determining the existence of differences among several population means. The case talks about

the textile manufacturer who has a large order to manufacture the cloth. The cloth is

manufactured using four different dying lines. It is given in the case that no line is used is for

more than one product but as the volume of order is large four lines are being used. The purpose

behind all this is to minimize the variance of the brightness of all the cloth produced. But later on

customer complaint that there are variance in the brightness. For this purpose ANOVA test of the

brightness of the cloth from four lines is conducted.

Hypothesis

H0: µ1=µ2=µ3=µ4

H1: Not all µ1, µ2, µ3 and µ4 are equal.

Where µ1, µ2, µ3 and µ4 are the average brightness of four different lines.

Solution

1. At 5% significant level,

ANOVA single factor

SUMMARY

Groups Count Sum Average Variance

line 1 16 1114.66 69.66625 5.910412

line 2 14 932.17 66.58357 6.422871

line 3 12 801.06 66.755 2.9703

line 4 14 1002.15 71.58214 7.812341

ANOVA

Source of Variation SS D f MS F cal P-value F crit

Between Groups 239.195 3 79.7316 13.532 1.20702E-06 2.7826

Within Groups 306.387 52 5.89206

Total 545.582 55

Interpretation:-

From the above table we can see that F calculated value is greater than F critical value. Therefore

we have to reject the null hypothesis. So we can say that company which are using different lines

for manufacturing cloth will be stopped.

From the manager point of view, here we can say that the company should improve the dying

quality for all the lines so they can manufacture better cloth and satisfy the customer wants and

needs.

2. To find significant difference in average brightness we have to first calculate the number

of combination. There are 6 different combination and calculated value in the form of

table is shown below:

H0 H1

µ1=µ2 µ1≠µ2

µ2=µ3 µ2≠µ3

µ3=µ4 µ3≠µ4

µ1=µ3 µ1≠µ3

µ1=µ4 µ1≠µ4

µ2=µ4 µ2≠µ4

Lines Mean Diff Significant diff

1&2 3.08267857 3.08

2&3 -0.1714286 0.17

3&4 -4.8271429 4.8

1&3 2.91125 2.91

1&4 -1.9158286 1.91

2&4 -0.1714286 4.99

K 4

N 56

N-K 52

qα 3.764

T 2.28

From the template, I found the T value i.e. 2.28. So by comparing T value with the significant

differences it can be said that pair 1&2, 3&4, 1&3 and 2&4 have significant differences in their

brightness. It suggests that if we are talking about pair 1&2, line 1 shows more average

brightness than line 2. So by improving the quality of line 1 we can reduce the variance of

brightness of the cloth manufactured. Similar is the case with other pairs.

3. From the above significant table that I have calculated it can be said that pair 2&4 have

more significant difference i.e. 5 (approx.) compared to other pairs in which line 4 has

greater average brightness than line 2. So we have to adjust the line 4 by reducing the

mean. The value of line 4 is: -

Mean = 71.58 – 2.28 = 69.3.

By doing this we can minimize the variance in brightness in all the cloth produced.

4. After comparing the significant difference of different combination with T it can be said

that line 1 and line 4 have more average brightness compared to other lines. But in the

above answer we have already adjust the line 4 so now we have to adjust the line 1 by

adjusting the mean. The value of line 1 is:-

Mean is 69.67 - 2.28 = 67.39 and as I mentioned in the above answer that line

4 is adjusted to 69.3. So from this result we can say that by adjusting line 1 and line 4 company

can produce better quality cloth.