Artificial Teeth: wear. In a study by J. Zeng et al., three materials for making artificial teeth-Endura, Duradent and Duracross-were tested for wear. Their results were published as the paper "In Vitro Wear Resistance of Three Types of Composite Resin Denture Teeth" (Journal of Prosthetic Dentistry, Vol. 94, Issue 5, pp. 453-457). Using a machine that stimulated grinding by two right first molars at 60strokes per minute for a total of 50,000strokes, the researchers measured the volume of material worn away, in cubic millimeters. Six pairs site of teeth were tested for each material. The data on the WeissStats site are based on the results obtained by the researchers. At the 5%significance level, do the data provide sufficient evidence to conclude that there is a difference in mean wear among the three materials?

Short Answer

Expert verified

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Step by step solution

01

Introduction

a.

Examine whether the figures provide enough evidence to prove that the three materials have distinct mean wear rates.

It is necessary to state the null and alternative hypotheses.

Null Hypothesis:

H0:μNortheast=μReduced=μControl=μEnlarged

That is, the data is insufficient to determine whether the three materials have distinct average wear rates.

Alternative hypothesis:

Ha: A minimum of one μ1in comparison to others.

That is, the data is sufficient to determine whether the three materials have distinct average wear rates.

The importance level here is, α=0.005

02

Computation

Calculate the value of the test statistic:

MINITAB procedures:

Step 1: Select Stat > ANOVA > One-Way Analysis from the menu bar.

Step 2:Enter the Volume column in the Response column.

Step 3: Enter the Material column in Factor.

Step 4: Press the OK key.

03

MINITAB output

One-way ANOVA on volume vs. material

Source DF SS MS F P

MATERIAL 2 2.02446 1.01223 341.16 0.000

Error 15 0.04450 0.00297

Total 17 2.06896

s=0.05447 R-Sq=97.85% R-Sq(adj)=97.56%

Level N Mean StDev

Duracross 6 0.22000 0.08000

Duradent 6 0.87000 0.03000

Endura 6 0.10998 0.04002

Pooled StDev =0.05447

The value of F is 341.16 and the p-value is 0.000, according to the MINITAB output.

04

MINITAB Output

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