Values of modulus of elasticity (MOE, the ratio of stress, i.e, force per unit a
ID: 3320957 • Letter: V
Question
Values of modulus of elasticity (MOE, the ratio of stress, i.e, force per unit area, to strain, .e., deformation per unit length, in GPa) and flexural strength (a measure of the ability to resist failure in bending, in MPa) were determined for a sample of concrete beams of a certain type, resulting in the following data: MOE 29.9 33.2 33.7 35.2 35.3 36.0 36.2 36.3 37.6 37.7 38.5 38.7 39.7 41.2 Strength 6.1 7.07.16.1 8.2 6.7 6.9 7.5 6.66.3 7.1 6.4 7.7 9.1 MOE 42.7 42.7 43.3 45.746.0 46.9 47.9 49.5 51.6 62.8 70.0 79.4 79.8 Strength 8.0 8.57.69.7 7.3 7.99.9 7.9 7.5 11.6 11.4 11.9 10.6 Fitting the simple linear regression model to the n 27 observations on xmodulus of elasticity and y flexural strength given in the data above resulted in y = 7.535, sf 0.185 when x = 40 and y = 9.738, sf 0.262 for x 60 (a) Explain why sf is larger when x = 60 than when x = 40 The closer x is to x, the smaller the value of sy The closer x is to y, the smaller the value of sy O The farther x is from y, the smaller the value of sy The farther x is from x, the smaller the value of sy (b) Calculate a confidence interval with a confidence level of 95% for the true average strength of all beams whose modulus of elasticity is 40 (Round your answers to three decimal places.) 23 13744x ) MPa (c) Calculate a prediction interval with a prediction level of 95% for the strength of a single beam whose modulus of elasticity is 40. (Round your answers to three decimal places.) MPa (d) If a 95% CI is calculated for true average strength when modulus of elasticity is 60, what will be the simultaneous confidence level for both this interval and the interval calculated in part (b)? The simultaneous confidence level for these intervals is at leastExplanation / Answer
Solutionb:
MOE <- c(29.9,33.2,33.7,35.2,35.3,36,36.2,36.3,37.6,37.7,38.5,38.7,39.7,41.2,42.7,
42.7,43.3,45.7,46,46.9,47.9,49.5,51.6,62.8,70,79.4,79.8)
Strength <- c(6.1,7,7.1,6.1,8.2,6.7,6.9,7.5,6.6,6.3,7.1,6.4,7.7,9.1,8,8.5,7.6,9.7,7.3,7.9,9.9,7.9,
7.5,11.6,11.4,11.9,10.6)
print(MOE)
print(Strength)
mod1.lm= lm(Strength~MOE)
newdata = data.frame(MOE=40)
predict(mod1.lm, newdata, interval="confidence",conf.level=0.95)
output:
fit lwr upr
1 7.53543 7.15368 7.91718
95% coonfidence interval is
7.154 to 7.917
7.154,7.917
Solutionc:
MOE <- c(29.9,33.2,33.7,35.2,35.3,36,36.2,36.3,37.6,37.7,38.5,38.7,39.7,41.2,42.7,
42.7,43.3,45.7,46,46.9,47.9,49.5,51.6,62.8,70,79.4,79.8)
Strength <- c(6.1,7,7.1,6.1,8.2,6.7,6.9,7.5,6.6,6.3,7.1,6.4,7.7,9.1,8,8.5,7.6,9.7,7.3,7.9,9.9,7.9,
7.5,11.6,11.4,11.9,10.6)
print(MOE)
print(Strength)
mod1.lm= lm(Strength~MOE)
newdata = data.frame(MOE=40)
predict(mod1.lm, newdata, interval="predict",conf.level=0.95)
95% prediction interval for MOE=40 is
5.649 to 9.422
5.649,9.422
fit lwr upr
1 7.53543 5.649014 9.421847
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