An article considered regressing y = 28-day standard-cured strength (psi) agains

Question

An article considered regressing

y = 28-day standard-cured strength (psi)

against

x = accelerated strength (psi).

Suppose the equation of the true regression line is

y = 1900 + 1.4x,

and that the standard deviation of the random deviation is 350 psi.

a. What is the probability that the observed value of 28-day strength will exceed 5000 psi when the value of accelerated strength is 2100? (Round your answer to four decimal places.)

b.

What is the probability that the observed value of 28-day strength will exceed 5000 psi when the value of accelerated strength is 2500? (Round your answer to four decimal places.)

c.

Consider making two independent observations on 28-day strength, the first for an accelerated strength of 2100and the second for

x = 2500.

What is the probability that the second observation will exceed the first by more than 1000 psi? (Round your answer to four decimal places.)

d.

Let

Y1

and

Y2

denote observations on 28-day strength when

x = x1

and

x = x2,

respectively. By how much would

x2

have to exceed

x1

in order that

P(Y2 > Y1) = 0.95?

(Round your answer to two decimal places.)

Explanation / Answer

Solution:

a) Regression value =1900+1.4x

the probability that the 28th day stregth will exceed 5000 psi when the accelerated strength x=2100

The predicted value is

y = 1900+1.4(2100)

=1900+2940

= 4840
calculate

p(Y exceeding 5000)

=1-p(Z exceeding 5000)

=1-p(Z exceeding (5000-4840)/350

=1-p(Z exceeding 0.4571)

For 0.4571 look at normal distribution table

=0.6772

=1-0.6772

= 0.3228

b) Y=1900+1.4x

x=2500

Y=1900+1.4×2500

=1900+3500

=5400

P(Y exceeding 5000)

=1-P(z exceecing 5000)

=1-p( z is 5000-5400)/350

=1-p(z at -1.1429)

=10.8729 = 0.1271

c)

Between 2100 and 2500

= 0.3228+0.1271
= 0.4499

= 44.99%

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