Answer to Question #284080 in Statistics and Probability for Leyman

Question

A product designer is testing the tensile strength of a new synthetic fiber that will be used to produce cloth for women’s jackets. Strength is usually affected by the percentage of cotton used in the blend of materials for the fiber. An experiment is conducted with five levels of cotton content and replicated the experiment five times. The data are shown in the following Table 2.

Table 2

Cotton Weight (%)

Observations

15

7

7

15

11

9

20

12

17

12

18

18

25

14

19

19

18

18

30

19

25

22

19

23

35

7

10

11

15

11

Source

Sum of Squares

Treatment

475.76

Error

161.2

Identify the following items in this study.

Treatments

Dependant variable

Number of replications

Construct an analysis of variance (ANOVA) table.

Is there evidence to support the claim that cotton content affects the mean tensile strength? Use α = 0.05

Use the Fisher LSD method to make comparisons between the pairs of means. What conclusions can you draw?

Answer 1

$a)$

The cotton weights are the treatments.

$b)$

The dependent variable is the tensile strength

$c)$

The number of replications is 5

$d)$

The ANOVA table for this analysis is given below.

Source. d.f. SS. MS F_c

Treatments 4 475.76 118.94 14.76

Error 20 161.20 8.06

Total 24 636.96

$e)$

The hypotheses tested are,

$H_0: \mu_1=\mu_2=\mu_3=\mu_4=\mu_5$

$against$

$H_1:$ At least two of the means are different.

The $F$ table value at $\alpha=0.05$ with 4 numerator degrees of freedom and 20 denominator degrees of freedom is,

$F_{0.05,4,20}=2.87$ and we reject the null hypothesis if $F_c\gt F_{0.05,4,20}$

Since $F_c=14.76\gt F_{0.05,4,20}=2.87$, the null hypothesis is rejected and we conclude that there is sufficient evidence to show that the cotton content affects the mean tensile strength at 5% significance level.

$f)$

To make comparisons between pairs of means, let us find the mean for each treatment level.

Treatment. Mean

15% cotton 9.8

20% cotton 15.4

25% cotton 17.6

30% cotton 21.6

35% cotton 10.8

The Fisher Least Significant difference formula is given by,

$LSD_{i,j}=t_{{\alpha\over 2},df}\sqrt{MSE({1\over n_i}+{1\over n_j})}$ ​for$i\gt j$ and$(i,j)$ is the pair to be compared.$t_{{\alpha\over 2},df}=t_{0.025,20}=2.085963$. $MSE=8.06$ is the mean square for error in the ANOVA table above. Therefore

$LSD_{i,j}=2.085963\sqrt{8.06({1\over5}+{1\over5})}=3.75(2dp)$

This value is common to all the possible pairs since all the groups share the same;

$a)$ ​significance level, $\alpha=0.5$ with degrees of freedom, $df=20$

$b)$ ​Mean square for error, $MSE=8.06$ and

$c)$ ​sample sizes for all levels $n=5$

Next is to compute the absolute difference in means for all possible pairs.

There are a total of $k={t(t-1)\over2}$ comparisons where $t=5$ is the number of treatment levels. Therefore, there are $k={5\times4\over 2}=10$ comparisons to be made.

$Pairs.$ ​ $i.$ $j.$ $|\bar{y}_i-\bar{y}_j|$

15% and 20% cotton 1 2 5.6

15% and 25% cotton 1 3 7.8

15% and 30% cotton 1 4 11.8

15% and 35% cotton 1 5 1

20% and 25% cotton 2 3 2.2

20% and 30% cotton 2 4 6.2

20% and 35% cotton 2 5 4.6

25% and 30% cotton 3 4 4

25% and 35% cotton 3 5 6.8

30% and 35% cotton 4 5 10.8

The hypotheses tested are,

$H_0:\mu_i=\mu_j \space vs \space H_1: \mu_i\not=\mu_j$

The null hypothesis of same means is rejected if $|\bar{y}_i-\bar{y}_j|\gt LSD_{i,j}$

We present the conclusions in the table below.

$Pairs.$ ​ $i.$ $j.$ $|\bar{y}_i-\bar{y}_j|$. $LSD_{i,j}$ $Significant?$

15% and 20% cotton 1 2 5.6 3.75 YES

15% and 25% cotton 1 3 7.8 3.75 YES

15% and 30% cotton 1 4 11.8 3.75 YES

15% and 35% cotton 1 5 1 3.75 NO

20% and 25% cotton 2 3 2.2 3.75 NO

20% and 30% cotton 2 4 6.2 3.75 YES

20% and 35% cotton 2 5 4.6 3.75 YES

25% and 30% cotton 3 4 4 3.75 YES

25% and 35% cotton 3 5 6.8 3.75 YES

30% and 35% cotton 4 5 10.8 3.75 YES

From the table above, difference in means for two pairs are insignificant. That is,

15% and 35% cotton and 20% and 25% cotton. The difference in means for the other pairs are significant at 5% significance level.