Values of modulus of elasticity (MOE, the ratio of stress, i.e., foroe per unit

Question

Values of modulus of elasticity (MOE, the ratio of stress, i.e., foroe per unit area, to strain, .e.,deformation per unit length, in GPa) and filexural strength (a measure of the ability to resist failure in bending, in MPa) were determined for a sample ot concrete beams of a certain type, resulting in the following data 29.9 33.0 33.6 35.3 35.4 35.9 36.0 36.1 37.5 37.5 38.6 39.0 39.5 41.0 MOE Strength 6.17.0 7.1 6.4 8.3 6.8 6.9 7.4 6.8 6.6 7.1 6.57.8 9.0 42.7 42.7 43.5 45.7 46.1 47.0 48.1 49.5 51.7 62.8 69.7 79.6 80.1 MOE Strength 8.08.9 7.7 9.6 7.37.5 9.5 8.0 7.6 116 11.5 11.8 10.6 Fitting the simple linear regression model to the n = 27 observations on x modulus of elasticity and y flexural strength given in the data above resulted in y = 7.581, sp = 0.177 when x = 40 and y = 9.721, S = 0.250 for x = 60 (a) Explain why is larger when x = 60 than when x = 40, The closer x is to x, the smaller the value of sp. The farther x is from y, the smaller the value of sy. The rather x is from x, the smaller the value of sp. The closer x is to y, the smaller the value of sp. (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.) MPa c calculate a prediction interval with a prediction level of 95% for the strength of a single beam wh se modulus of elasticity is 40 Round your ans ers to th ee 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 simultaneaus confidence level for these intervals is at least

Explanation / Answer

a) option A) is correct

b)

predict ( mod, data.frame(x=40) ,interval = "confidence")
    fit           lwr    upr
1   7.581192   7.216713 7.945672

(7.217 , 7.946)

c)

predict ( mod, data.frame(x=40) ,interval = "prediction")
     fit                      lwr           upr
1      7.581192     5.779183    9.383202

(5.779 , 9.383)

d)

predict ( mod, data.frame(x=60) ,interval = "confidence")
     fit               lwr           upr
1     9.72051     9.205454     10.23557

0%

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