My Notes Ask Your Teacher An article considered regressing y=28-day standard-cur

Question

My Notes Ask Your Teacher An article considered regressing y=28-day standard-cured strength (psi) against x - accelerated strength (psl). Suppose the equation of the true regression line is y 1900+ 1.3x, and that the standard deviation of the random deviation c 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 2600? (Round your answer to four decimal places.) (e) Consider making two independent observations on 28-day strength, the first for an accelerated strength of 2100 and the second for x 2600. What is the probability that the second observation will exceed the first by more than 1000 psi? (Round your answer to four decimal places.) Let respectively. By how much would X2 have to exceed in order that Pf½ > r»-0.95? (Round your answer to two decimal places.) (d) and ½ denote observations on 28-day strength when x=#1 and x=x2. You may need to use the appropriate table in the Appedix of Tables to answer this question Need Help?ReadITalk to a Tutor

Explanation / Answer

Regression value =1900+1.3x

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

The predicted value is

y=1900+1.3(2100)

=1900+2730

=4630

calculate

p(Y exceeding 5000)

=1-p(Z exceeding 5000)

=1-p(Z exceeding (5000-4630)/350

=1-p(Zexceeding1.0571)

For 1.0571 look at normal distribution table

=0.8531

=1-0.8531

=0.1469

b. Y=1900+1.3x

x=2600

Y=1900+1.3×2600

=1900+3480

=5280

P(Y exceeding 5000)

=1-P(z exceecing 5000)

=1-p( z is 5000-5389)/350

=1-p(z at -1.11142)

=1-(0.8665)

=0.1335.

c.

Between 2100 and 2600

0.1464+0.1335

=.2799

=27.99%

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