An experiment investigated the weakening effect of cleaning treatments on the br

Question

An experiment investigated the weakening effect of cleaning treatments on the breaking strength of strands taken from of a 1/2-inch climbing rope. The strands, about 30 inches long each, were subjected to 5 treatments:

• No treatment (control).

• Soaked in water for 66 hours.

• Soaked in full-strength household bleach for 66 hours.

• Soaked in acetone for 72 hours.

• Soaked in denatured alcohol for 72 hours.

After removal from their respective baths, the samples were air dried for 24 hours. The bleach sample was rinsed in water before air drying. 3 strands were subjected to each treatment, so there were 15 samples to break by pulling between two 1/2-inch steel pins. Here are the results: treatment breaking strength (pounds) nothing 48 50 54 water 50 47 52 bleach 48 52 46 acetone 50 48 53 alcohol 46 48 44

(a) Make a preliminary graph of the data. Why did you choose the graph that you did and what does it tell you?

(b) Create an ANOVA table using R.

(c) Evaluate the ANOVA assumptions graphically. Was ANOVA appropriate here?

(d) Based on the ANOVA table, make a conclusion in the context of the problem

Explanation / Answer

Here there are five groups.

Here we have to test the hypothesis that,

H0 : All the population means are equal.

H1 : Atleast one of the mean is differ than 0.

We have to make ANOVA in R.

> y1=c(48,50,54)
> y2=c(50,47,52)
> y3=c(48,52,46)
> y4=c(50,48,53)
> y5=c(46,48,44)
> y=c(y1,y2,y3,y4,y5)
> n=rep(3,5)
> n
[1] 3 3 3 3 3
> group=rep(1:5,n)
> group
[1] 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5
> data=data.frame(y=y,group=factor(group))
> fit=lm(y~group,data)
> anova(fit)
Analysis of Variance Table

Response: y
Df Sum Sq Mean Sq F value Pr(>F)
group 4 42.267 10.5667 1.4953 0.2754
Residuals 10 70.667 7.0667   
> df=anova(fit)[,"Df"]
> names(df)=c("trt","err")
> df
trt err
4 10
> alpha=c(0.05,0.01)
> qf(alpha,df["trt"],df["err"],lower.tail=FALSE)
[1] 3.478050 5.994339

Here we see that P-value = 0.2754

Assume alpha = level of significance = 5% = 0.05

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : All the population means are equal.

To use the ANOVA test we made the following assumptions:

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