Consider the accompanying data on flexural strength (MPa) for concrete beams of
ID: 3052368 • Letter: C
Question
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type 5.0 7.2 7.3 6.3 8.1 6.8 7.0 7.0 6.8 6.5 7.0 6.3 7.9 9.0 8.6 8.7 7.8 9.7 7.4 7.7 9.7 7.9 7.7 11.6 11.3 11.8 10.7 The data below give accompanying strength observations for cylinders. 6.4 5.8 7.8 7.1 7.2 9.2 6.6 8.3 7.0 8.1 7.2 8.1 7.4 8.5 8.9 9.8 9.7 14.1 12.6 12.0 Prior to obtaining data, denote the beam strengths by X1, ···, xm and the cylinder strengths by Y1, ·.. , Yn. Suppose that the X,'s constitute a random sample from a distribution with mean 1 and standard deviation 1 and that the Y's form a random sample (independent of the X's) from another distribution with mean 2 and standard deviation 2 (a) Use rules of expected value to show that X-Y is an unbiased estimator of 1-12 nm Calculate the estimate for the given data. (Round your answer to three decimal places.) 486 MPa (b) Use rules of variance to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a). Compute the estimated standard error. (Round your answer to three decimal places.) 587 MPa (c) Calculate a point estimate of the ratio /2 of the two standard deviations. (Round your answer to three decimal places.) 798 (d) Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference X-Y between beam strength and cylinder strength. (Round your answer to two decimal places.) 12 XMPaExplanation / Answer
answer:
a) we know x has mean µ1 and var is 12 and y has mean µ2 and var is 22.
Therefore xbar has mean and var are same as x .i.e. µ1 and 12 .
Symmilarly ybar has mean and var are µ2 and 22. respectivly.
therefore E (xbar-ybar)= E(xbar)-E(ybar)
=µ1- µ2
R output for calulating estimate is given below.
> x=c(5,7.2,7.3,6.3,8.1,6.8,7,7,6.8,6.5,7,6.3,7.9,9,8.6,8.7,7.8,9.7,7.4,7.7,9.7,7.9,7.7,11.6,11.3,11.8,10.7)
> y=c(6.4,5.8,7.8,7.1,7.2,9.2,6.6,8.3,7,8.1,7.2,8.1,7.4,8.5,8.9,9.8,9.7,14.1,12.6,12)
> xbar=mean(x)
> ybar=mean(y)
> est=xbar-ybar;est
[1] -0.486
b)
here x and y are independent. therefore xbar and ybar also independent.
Var(xbar- yabr) = var(xbar) + var(ybar) -2 cov(xbar, ybar)
= var(xbar)+ var(ybar) ............... cov=0
= 12 /n1 + 22 /n2
sd(xbar- ybar) =sqrt(var(xbar-ybar))
= sqrt(12 /n1 + 22 /n2)
R output is given below for calculating estimate:
> n1=length(x)
> n2=length(y)
> v1=var(x)/n1
> v2=var(y)/n2
> se=sqrt(v1+v2)
> se
[1] 0.587