Question

Suppose the value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations:

116.4

115.7

114.7

115.1

115.5

(a) Calculate x.
GPa

Calculate the deviations from the mean. (Enter your answers to two decimal places.)

x	116.4	115.7	114.7	115.1	115.5
deviation

(b) Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation. (Round your answers to three decimal places.)

s2	=	GPa2
s	=	GPa

(c) Calculate s² by using the computational formula for the numerator S_xx. (Round your answer to three decimal places.)
GPa²

(d) Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of these transformed values. (Round your answer to three decimal places.)
GPa²

Compare it to s² for the original data.

The variance in part (d) is greater than the variance in part (b).The variance in part (d) is equal to the variance in part (b).    The variance in part (d) is smaller than the variance in part (b).

suppose the value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations: 16.4 115.7 114.7 115.1 115.5 a) Calculate x. Gra Calculate the deviations from the mean. Enter your answers to two decimal places.) 115.5 11G.4 115.7 deviation (b) use tha deviations calculated in part (a) to ootain tha sample variance and th sample standard deviation. (Round your answers to three dacimal places.) GPa 2 GPa (e) Calculate s2 by using the computational formula for the numerator S (Round your answer to three decimal places.) Cna2 (d) Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of thesc transformed values. (Round your answer to three decimal places.) cpa2 Campare it to s tor tha original data. The variance in part (d) greater than the varianca in part (b). 春The variance in part (d) i"qual to the varianca in part (b). ·The variance in part (d) smaller than the variance in part (b).

Answer 1

a) = (116.4 + 115.7 + 114.7 + 115.1 + 115.5)/5 = 115.48

x                         116.4            115.7               114.7                 115.1              115.5

deviation                0.92             0.22               -0.78                 -0.38         0.02       

b) s^2 = ((0.92)^2 + (0.22)^2 + (-0.78)^2 + (-0.38)^2 + (0.02)^2)/4 = 0.412

    s = = 0.642

у/0.412

c) = 116.4 + 115.7 + 114.7 + 115.1 + 115.5 = 577.4

    = (116.4)^2 + (115.7)^2 + (114.7)^2 + (115.1)^2 + (115.5)^2 = 66679.8

     

2-1577412 66679.8 =

            = 0.412

d) By subtracting 100 from each observation, the new sample will be

16.4, 15.7, 14.7, 15.1, 15.5

= (16.4 + 15.7 + 14.7 + 15.1 + 15.5)/5 = 15.48

s^2 = ((16.4 - 15.48)^2 + (15.7 - 15.48)^2 + (14.7 - 15.48)^2 + (15.1 - 15.48)^2 + (15.5 - 15.48)^2)/4 = 0.412

The variance in part (d) is equal to the variance in part(b).