Question

**MINITAB IS required for this exercise also**

The printing department of a publishing company wants to determine whether there are differences in durability between three types of book bindings. Twenty-five books with each type of binding were selected and placed in machines that continually opened and closed them. The numbers of openings and closings until the pages separated from the binding were recorded.

a. What techniques should be considered to determine whether differences exist between the types of bindings? What are the required conditions? How do you decide which technique to use?

b. If we know that the number of openings and closings is not normally distributed, test to determine whether differences exist between the types of bindings.

Binding 1   Binding 2   Binding 3
2550   3024   2671.0
3226   2725   2905.0
2976   3655   3218.0
2633   3034   3136.0
3242   2897   2891.0
2674   3189   3449.0
2599   3269   3427.0
2744   3045   2767.0
3493   3157   3385.0
2844   2994   2711.0
3313   2774   3019.0
2742   2762   3087.0
3145   3454   3396.0
3090   2849   2531.0
3248   3318   3311.0
3492   3635   2840.0
2560   3022   2507.0
3130   3350   3081.0
2840   3242   3323.0
2935   3404   3350.0
3050   3306   3058.0
2991   2906   3462.0
3143   3205   2955.0
3118   3522   3220.0
3377   3536   2569.0

Answer 1

a) Since here three different individual samples we cannot perform t-test for difference of means. If more than two individual samples are there then we have to perform one way ANOVA.

Assumptions to be met:-

i) All samples collected are random and individual from one another.

ii) All observations in Samples are randomly selected individual from one another.

iii) Test for homogenity of variances

Test for Equal Variances: Binding 1, Binding 2, Binding 3

Method

Null hypothesis	All variances are equal
Alternative hypothesis	At least one variance is different
Significance level	α = 0.05

95% Bonferroni Confidence Intervals for Standard Deviations

Sample	N	StDev	CI
Binding 1	25	285.998	(227.048, 398.404)
Binding 2	25	274.706	(217.019, 384.552)
Binding 3	25	305.313	(241.867, 426.215)

Individual confidence level = 98.3333%

Tests

Method	Test Statistic	P-Value
Multiple comparisons	—	0.770
Levene	0.17	0.846

Test for Equal Variances: Binding 1, Binding 2, Binding 3

the p-value is more than 0.05 hence we fail to reject null hypothesis and conclude that the variances are equal.

One-way ANOVA: Binding 1, Binding 2, Binding 3

Method

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal
Significance level	α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor	Levels	Values
Factor	3	Binding 1, Binding 2, Binding 3

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Factor	2	363162	181581	2.17	0.121
Error	72	6011382	83491
Total	74	6374544

Model Summary

S	R-sq	R-sq(adj)	R-sq(pred)
288.949	5.70%	3.08%	0.00%

Means

Factor	N	Mean	StDev	95% CI
Binding 1	25	3006.2	286.0	(2891.0, 3121.4)
Binding 2	25	3171.0	274.7	(3055.8, 3286.2)
Binding 3	25	3050.8	305.3	(2935.6, 3166.0)

Pooled StDev = 288.949

Interval Plot of Binding 1, Binding 2, ...

b) If we know that the number of openings and closings is not normally distributed, test to determine whether differences exist between the types of bindings. We have to perform non parametric tests. Kurshal wallis test

Kruskal-Wallis Test: Response versus Binding

Descriptive Statistics

Binding	N	Median	Mean Rank	Z-Value
1	25	3050	33.1	-1.38
2	25	3189	44.4	1.80
3	25	3081	36.5	-0.42
Overall	75		38.0

Test

Null hypothesis	H₀: All medians are equal
Alternative hypothesis	H₁: At least one median is different

Method	DF	H-Value	P-Value
Not adjusted for ties	2	3.55	0.170
Adjusted for ties	2	3.55	0.170

p-value is more than 0.05 hence we fail to reject null hypothesis and conclude that all medians are equal.