Question

Show work please.

An article in the Tappi Journal (March, 1986) presented data on green liquor NazS concentration (in grams per liter) and paper machine production (in shown as follows y 40 42 49 46 44 48 46 43 53 52 54 57 58 825 830 890 895 890 910 915 960 990 1010 1012 1030 1050 Assume y green liquor Na2S concentration and x- production in a paper mil Round your answers to 3 decimal places. Find a 99.00% confidence interval on each of the following: (a) f, (b) β) (c) Mean NaS concentration when productionx876 tons/day (d) Find a 99.00% prediction interval on Na2S concentration when x-876 torsday.

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
825	40	12996.00	74.225	982.154
830	42	11881.00	43.763	721.077
890	49	2401.00	0.148	-18.846
895	46	1936.00	6.840	115.077
890	44	2401.00	21.302	226.154
910	48	841.00	0.379	17.846
915	46	576.00	6.840	62.769
960	43	441.00	31.533	-117.923
990	53	2601.000	19.225	223.615
1010	52	5041.000	11.456	240.308
1012	54	5329.000	28.994	393.077
1030	57	8281.000	70.302	763.000
1050	58	12321.000	88.071	1041.692

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	12207	632	67046.000	403.1	4650
mean	939.00	48.62	SSxx	SSyy	SSxy

sample size ,   n =   13      
here, x̅ = Σx / n=   939.00   ,     ȳ = Σy/n =   48.62
              
SSxx =    Σ(x-x̅)² =    67046.000000      
SSxy=   Σ(x-x̅)(y-ȳ) =   4650.0     

estimated slope , ß1 = SSxy/SSxx =   4650.0   /   67046.000   =   0.0694
                  
intercept,   ß0 = y̅-ß1* x̄ =   -16.5093          
                  
so, regression line is   Ŷ =   -16.5093   +   0.069   *x

SSE=    (SSxx * SSyy - SS²xy)/SSxx =     80.574
        
std error ,Se =     √(SSE/(n-2)) =     2.70646

a)

confidence interval for intercept  

α=   0.01
DF=n-2 = 11   
t critical value=t α/2 =    3.1058   [excel function: =t.inv.2t(α/2,df) ]          
estimated std error of slope =Se(ßo) =    Se*√(1/n+x̅²/Sxx)=       9.8435      
margin of error ,E=   t*std error =        30.5719      
                  
lower confidence limit = ß̂o - E =   -16.5093   -   30.5719   =   -47.081
upper confidence limit= ß̂o + E =   -16.5093   +   30.5719   =   14.063

b)

confidence interval for slope                  
α=   0.01              
t critical value=   t α/2 =    3.106   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    2.70646   /√   67046.00   =   0.010
                  
margin of error ,E= t*std error =    3.106   *   0.010   =   0.032
estimated slope , ß^ =    0.0694              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.0694   -   0.032   =   0.037
upper confidence limit=estimated slope + margin of error =   0.0694   +   0.032   =   0.102

c)

Predicted Y at X=   876   is                  
Ŷ =   -16.509   +   0.069   *   876   =   44.246

d)

X Value=   876                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   13                      
Degrees of Freedom,df=n-2 =   11                      
critical t Value=tα/2 =   2.201   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    939.00                      
Σ(x-x̅)² =Sxx   67046.0                      
Standard Error of the Estimate,Se=   2.71                      
                          
Predicted Y at X=   876   is                  
Ŷ =   -16.509   +   0.069   *   876   =   44.246
                          

                          
For Individual Response Y                          
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   2.8848                      
margin of error,E=t*std error=t*S(ŷ)=    2.2010   *   2.88   =   6.3494      
                          
Prediction Interval Lower Limit=Ŷ -E =   44.246   -   6.3494   =   37.897      
Prediction Interval Upper Limit=Ŷ +E =   44.246   +   6.3494   =   50.595