Question

As concrete cures, it gains strength. The following data represent the 7-day and 28-day strength in pounds per square inch (psi) of a certain type of concrete. Complete parts (a) through (e) below.

7-Day Strength (psi), x	3380	2480	3330	2890	3390
28-Day Strength (psi), y	5020	4120	4850	4620	5220

1. Treating the 7-day strength as the explanatory variable, x, use technology to determine the estimates of β0 and β1.

β0 ≈ b0 =

(Round to one decimal place as needed.)

β1 ≈ b1 =

(Round to four decimal places as needed.)

2. Compute the standard error of the estimate, se.

Se =

(Round to one decimal place as needed.)

3. A normal probability plot suggests that the residuals are normally distributed. Determine Sb1. Use the answer for part b.

Sb1 =

(Round to four decimal places as needed.)

4. A normal probability plot suggests that the residuals are normally distributed. Test whether a linear relation exists between 7-day strength and 28-day strength at the a = 0.05 level of significance.

• State the null and alternative hypotheses.

• Determine the P-value of this hypothesis test.

P value =

(Round to three decimal places as needed.)

• What is the conclusion drawn?

Reject H0 and conclude that linear relation does not exist.

Reject H0 and conclude that linear relation does exist.

Do Not Reject H0 and conclude that linear relation does not exist.

Do Not Reject H0 and conclude that linear relation does exist.

5. Construct a 95% confidence interval about the slope of the true least-squares regression line.

Lower bound=

(Round to three decimal places as needed.)

Upper bound =

(Round to three decimal places as needed.)

6. What is the estimated mean 28-day strength of this concrete if the 7-day strength is 3000psi?

A good estimate of the mean of 28-day strength is ___ psi.

(Round to two decimal places as needed.)

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
3380	5020	81796.00	64516.0	72644.0
2480	4120	376996.00	417316.0	396644.0
3330	4850	55696.00	7056.0	19824.0
2890	4620	41616.00	21316.0	29784.0
3390	5220	87616.00	206116.0	134384.0

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	15470	23830	643720.000	716320.0	653280
mean	3094.000	4766.00	SSxx	SSyy	SSxy

sample size ,n    =5

here,   x̅   =3094.000,ȳ =4766.0000

SSxx = Σ(x-x̅)² = 643720.00

SSxy=Σ(x-x̅)(y-ȳ) =653280.0

a)

slope , ß1 = SSxy/SSxx =1.0149

intercept,ß0 = y̅-ß1* x̄ =1626.1

b)

SSE=(Sx*Sy - S²xy)/Sx = 53338.02

std error ,Se = √(SSE/(n-2)) = 133.3

c)

estimated std error of slope =Se(ß1) = s/√Sxx = 33.3/√643720 = 0.1662

d)

slope hypothesis test

Ho:ß1=0

H1:ß1╪0

n=5

alpha=0.05

estimated std error of slope =Se(ß1) = 0.1662

  

t stat = ß1 /Se(ß1) = 6.107

  

  

p-value = 0.009 [excel function: =t.dist.2t(6.107,3) ]

p-value <α=0.05, reject Ho

decision : p-value<α , reject Ho   

Reject H0 and conclude that linear relation does exist.

e)

α=0.05,

df=n-2 = 3

t critical value=t α/2 = 3.1824 [excel function: =t.inv.2t(0.05/2,3) ]

margin of error ,E=t*std error = 0.5289

lower confidence limit = ß̂1-E =0.486

upper confidence limit=ß̂1+E =1.544

f)

Ŷ    =1626.1+1.0149*x

here, x=28

so, Ŷ    =1626.0505+1.0149*28 = 1654.47(answer)