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 (f) below. 7-Day Strength (psi), x:3340 3380 2300 2890 2480 28-Day Strength (psi),y: 4630 5020 4070 4620 4120 Treating the 7-day strength as the explanatory variable, x, use technology to determine the estimates of β0 and β1. 0β≈b0= β1≈b1= P-Value: What is the 95% confidence interval about the slope of the true least-squares regression line? Upper/lower bound. Compute the standard error of the estimate, se= A normal probability plot suggests that the residuals are normally distributed. Determine sb1. sb1= 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 α=0.05 level of significance. State the null and alternative hypotheses. Reject or do not reject? What is the estimated mean 28-day strength of this concrete if the 7-day strength is 3000 psi?

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
3340	4630	213444.0000	19044.0000	63756.0000
3380	5020	252004.0000	278784.0000	265056.0000
2300	4070	334084.0000	178084.0000	243916.0000
2890	4620	144.0000	16384.0000	1536.0000
2480	4120	158404.0000	138384.0000	148056.0000

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	14390.00	22460.00	958080.00	630680.00	722320.00
mean	2878.00	4492.00	SSxx	SSyy	SSxy

sample size ,   n =   5          
here, x̅ = Σx / n=   2878.000   ,     ȳ = Σy/n =   4492.000  
                  
SSxx =    Σ(x-x̅)² =    958080.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   722320.0          
                  
estimated slope , ß1 = SSxy/SSxx =   722320.0   /   958080.000   =   0.75392
                  
intercept,   ß0 = y̅-ß1* x̄ =   2322.20524          

alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    169.416   /√   958080.00   =   0.1731
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.7539   /   0.1731   =   4.3559
                  
t-critical value=    3.182   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   3              
p-value =    0.022358              
-----------

confidence interval for slope                  
α=   0.05              
t critical value=   t α/2 =    3.182   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    169.41590   /√   958080.00   =   0.173
                  
margin of error ,E= t*std error =    3.182   *   0.173   =   0.551
estimated slope , ß^ =    0.7539              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.7539   -   0.551   =   0.203
upper confidence limit=estimated slope + margin of error =   0.7539   +   0.551   =   1.305

---------

std error ,Se =    √(SSE/(n-2)) =    169.4159
----------

estimated std error of slope,Sb1 = Se/√Sxx =    169.41590   /√   958080.00   =   0.173

---------
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   5              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    169.416   /√   958080.00   =   0.1731
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.7539   /   0.1731   =   4.3559
                  
t-critical value=    3.182   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   3              
p-value =    0.022358              
decison :    p-value<α , reject Ho              

--------

Predicted Y at X=   3000   is          
Ŷ =   2322.2052   +   0.7539   *3000=   4583.9788