As concrete cures, it obviously starts to gain strength. The data in the table b
ID: 3322141 • Letter: A
Question
As concrete cures, it obviously starts to gain strength. The data in the table below represent the 7-day and correlation between the 7-day and 28-day concrete strengt represent using these two variables is summarized as , Bo+ Px, in which the betas are your coefficients pounds per square inch (psi) Suppose you are try ing to show that there is a definitive h. The linear regression model you are hoping to found by running a LinRegT Test. a) State the null and alternate hypothesis below using both symbols and words: Ho: L1 7-day Strength L2 28-day Strength 3390 2430 2890 3330 2480 3380 2660 2620 3340 5220 4620 4120 5020 4890 4630 b) Based on your LinRegT Test, what do you find your P-value to be? c) How would you interpret this result in simple language (oriented around your hypotheses)? d) Now, calculate a 95% prediction interval at x-3200 psi (7-day Strength). You can build a specific prediction interval, using x and the prediction interval formula shown below. You will probably want to place (-and Oi-into L3 and L4 respectively in order to summalc or simply modify your Challenger spreadsheet to make this calculation easier. The formulas are provided just in case you need to know the mathematical connections. Your prediction interval for 28-day Strength, based on x 3200 psi: ( ) , A level 1001-a)% prediction interval for the quantity A+ Ax is given by (7.44) where speed = n/i +-+ e) What is the R2 coefficient of determination (rounded to two decimals) and state how would you could interpret its meaning in the context of concrete strength?Explanation / Answer
R codes :
> x=c(2300,3390,2430,2890,3330,2480,3380,2660,2620,3340)
> y=c(4070,5220,4640,4620,4850,4120,5020,4890,4190,4630)
> model=lm(y~x)
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-304.80 -215.14 -44.22 203.07 415.16
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2675.5619 598.4958 4.470 0.00208 **
x 0.6764 0.2055 3.291 0.01100 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 271 on 8 degrees of freedom
Multiple R-squared: 0.5752, Adjusted R-squared: 0.5221
F-statistic: 10.83 on 1 and 8 DF, p-value: 0.011
> predict(model,newdata=data.frame(x=3200),interval="predict")
fit lwr upr
1 4840.101 4167.467 5512.735
a)Null hypothesis -> H0 : beta0 = beta1 = 0.
Alternative hypothesis -> H1 : not H0
b) p-value = 0.011
c) Since p-value < 0.05, we reject H0 in favour of H1 at 5% level and conclude that the coefficients of regression are significant and the regression significantly helps in predicting y.
d)Prediction interval for x = 3200 : (4167.467,5512.735)
e) R2 = 0.57 which means approx. 57% of the total variation is explained by the fitted regression model.
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