Here are the graph and table that I got from inputting the data they talked abou

Question

Here are the graph and table that I got from inputting the data they talked about in Problem 12 into JMP. I already did part A by getting the graph and tables below just need help with parts b-g. Please and Thank You!

12. Use sm read. table("Strength MoE.txt", header-T) to read the data on cement's modulus of elasticity and strength into the R data frame sm. Copy the data into the R objects x and y by x-smSMoE and y-smSStrength, and use R commands to complete the following. a. the regression ANOVA table, b. r c. Bo and along with their standard errors, default tstatistics and the p-values associated with the t statistics, d. 95 percent confidence intervals for BI, e. 95 percent confidence intervals for the mean responses when i) modulus of elasticity is 60 in problem 12 f 95 percent prediction intervals for the responses when i) modulus of elasticity is 60 in prob- lem 12 a g a brief interpretation of these results

Explanation / Answer

Solutionb:

r2 is called coefficient of determination here it is

r2=0.0177 from R output i attached

1.77% variation in dependent variable is explained by the model.

good model

Kindly provide the regression output to answer remaining questions

is it the dats:

Call:
lm(formula = strength ~ MoE, data = MOE)

Residuals:
Min 1Q Median 3Q Max
-7.999 -5.591 -0.642 0.707 56.517

Coefficients:
Estimate Std. Error t value
(Intercept) 18.4052 9.2400 1.992
MoE -0.1083 0.1610 -0.673
Pr(>|t|)
(Intercept) 0.0574 .
MoE 0.5074
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Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 12.05 on 25 degrees of freedom
Multiple R-squared: 0.01777,   Adjusted R-squared: -0.02151
F-statistic: 0.4524 on 1 and 25 DF, p-value: 0.5074

Regression eq is

strength=18.4052-0.1083(MoE)

slope=-0.1083

y intercept=18.4052

Solutionb:

y intercept bet=18.4052

slope =beta1=-0.1083

std error beta0=9.2400

t (betao)=1.993=2

std erro beta1=0.1610

t(slope)=-0.673

p value(beta0)=0.0574

p value(beta1)=0.5074

Solutiond

output form excel:

95% confidence interval for slope=

Solutionf:

when x=MOE=60

y=18.40516-0.10832(60)=11.90596

attach(MOE)
m2.lm= lm(strength ~ MoE)
newdata = data.frame(MoE=60)
View(newdata)
predict(m2.lm,newdata, interval="predict")

ouput:

fit lwr upr
1 11.90599 -13.41994 37.23193

lower prediction limit=-13.419

upper prediction limit=37.232

Solutione:

predict(m2.lm,newdata, interval="confidence")
fit lwr upr
1 11.90599 6.904126 16.90786

lower limtit=6.904126

upper limti=16.907

MoE strength 34.36 71.2 46.28 10.35 49.25 8.89 37.23 7.85 31.63 6.98 70.65 11.5 65.29 11.36 57.27 11.56 64.45 12.09 78.52 13.93 67.45 11.5 63.8 11.79 44.23 7.68 40.55 7.25 74.75 11.6 34.7 6.94 70.81 11.55 51.3 9.71 71.99 10.93 47.54 9.82 51.15 9.88 43.86 8.15 76.77 13.25 65.95 11.36 71.42 12.12 46.54 7.07 41.8 8.2

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