ENS6152 Steel Design Assignment Help
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From the above plot, it can be identified that:
The pH value of around 6.5 is the most prevalent in the different lakes irrespective of their sizes.
For smaller lakes, the pH value is lying to acidic side (i.e., having a pH value less than 7), while the lakes that are bigger are both acidic as well as basic.
The largest lake is the one at the highest pH value having an exceptionally high value for a sigficantly larger area as compared to the other lakes.
For using the principal of least squares for fitting a regression line to the data where pH is the dependent variable (y) and area is the independent variable (x), following calculation is needed to be done:
Observation 
Area (x) 
ph (y) 
xy 
x^{2} 
y^{2} 
1 
33 
6.6 
217.8 
1089 
43.56 
2 
161 
6.4 
1030.4 
25921 
40.96 
3 
189 
6.5 
1228.5 
35721 
42.25 
4 
149 
6.9 
1028.1 
22201 
47.61 
5 
47 
7.1 
333.7 
2209 
50.41 
6 
170 
7.5 
1275 
28900 
56.25 
7 
352 
8.8 
3097.6 
123904 
77.44 
8 
187 
6.4 
1196.8 
34969 
40.96 
9 
76 
5.9 
448.4 
5776 
34.81 
10 
52 
6.7 
348.4 
2704 
44.89 
11 
175 
7.1 
1242.5 
30625 
50.41 
12 
53 
6.6 
349.8 
2809 
43.56 
13 
200 
8.0 
1600 
40000 
64 
Sum 
1844 
90.5 
13397 
356828 
637.11 
From the above table, Σx = 1844, Σy = 90.5, Σxy = 13397, Σx^{2} = 356828, Σy^{2} = 637.11 n is the sample size (13, in our case).
We will use the following formula for finding a and b, where the regression line is y = a + bx
a= 6.13, b= 0.006
Hence the regression line is y = 6.127 + 0.005x
For the ANOVA test, following table is needed to be calculated:
DF 
SS 
MS 
F 

Regression 
1 
SSR= ∑ ( 
MSR=SSR/1 
F^{∗}=MSR/MSE 
Residual 
n2 
SSE=∑(yi− 
MSE=SSE/(n−2) 

Total 
n1 
SSTO=∑(y_{i}− 

Where
Following is the calculation for the table
Observation 
Area (x) 
ph (y) 
xy 
x^2 
y^2 
estimated y 
(yestimated y)^2 
(estimated y avg. of y)^2 
(yavg. of y)^2 
1 
33 
6.6 
217.8 
1089 
43.56 
6.32 
0.08 
0.41 
0.13 
2 
161 
6.4 
1030.4 
25921 
40.96 
7.07 
0.46 
0.01 
0.32 
3 
189 
6.5 
1228.5 
35721 
42.25 
7.24 
0.55 
0.08 
0.21 
4 
149 
6.9 
1028.1 
22201 
47.61 
7.00 
0.01 
0.00 
0.00 
5 
47 
7.1 
333.7 
2209 
50.41 
6.40 
0.48 
0.31 
0.02 
6 
170 
7.5 
1275 
28900 
56.25 
7.13 
0.14 
0.03 
0.29 
7 
352 
8.8 
3097.6 
123904 
77.44 
8.20 
0.36 
1.53 
3.38 
8 
187 
6.4 
1196.8 
34969 
40.96 
7.23 
0.68 
0.07 
0.32 
9 
76 
5.9 
448.4 
5776 
34.81 
6.57 
0.46 
0.15 
1.13 
10 
52 
6.7 
348.4 
2704 
44.89 
6.43 
0.07 
0.28 
0.07 
11 
175 
7.1 
1242.5 
30625 
50.41 
7.16 
0.00 
0.04 
0.02 
12 
53 
6.6 
349.8 
2809 
43.56 
6.44 
0.03 
0.27 
0.13 
13 
200 
8.0 
1600 
40000 
64 
7.30 
0.48 
0.12 
1.08 
Sum 
1844 
90.5 
13397 
356828 
637.11 
90.51 
3.80 
3.29 
7.09 
Following is the ANOVA table:

DF 
SS 
MS 
F 
Regression 
1 
3.291011 
3.291011 
9.527217 
Residual 
11 
3.799758 
0.345433 

Total 
12 
7.090769 


We are testing the null hypothesis H_{0}: b = 0 against the alternative hypothesis Ha: b ≠ 0.
If b = 0 then F = 1
If F > 1 then b ≠ 0, which means there is a linear relationship
In view of the fact there F = 9.52, we can say there is a linear relationship
Following are the assumptions of Regression
linearity and additivity – This can be seen from normal probability plot
statistical independence – This can be seen from the Versus fits
homoscedasticity  This can be seen from the Versus Orders
normality  This can be seen from the histogram
The regression line is y = 6.128 + 0.00588x
If x = 2050 then pH = 18.18
For developing a 99% CI for this prediction, following is the computation needed:
Computing the values, following is the confidence interval (6.88, 29.47)
For using the principal of least squares for fitting a regression line to the data where pressure is the dependent variable (y) and steam is the independent variable (x), following calculation is needed to be done:
Observation 
steam (x) 
pressure (y) 
xy 
x^2 
y^2 
1 
35.3 
10.98 
387.59 
1,246.09 
120.56 
2 
29.7 
11.13 
330.56 
882.09 
123.88 
3 
30.8 
12.51 
385.31 
948.64 
156.50 
4 
58.8 
8.4 
493.92 
3,457.44 
70.56 
5 
61.4 
9.27 
569.18 
3,769.96 
85.93 
6 
71.3 
8.73 
622.45 
5,083.69 
76.21 
7 
74.4 
6.36 
473.18 
5,535.36 
40.45 
8 
76.7 
8.5 
651.95 
5,882.89 
72.25 
9 
70.7 
7.82 
552.87 
4,998.49 
61.15 
10 
57.5 
9.14 
525.55 
3,306.25 
83.54 
11 
46.4 
8.24 
382.34 
2,152.96 
67.90 
12 
28.9 
12.19 
352.29 
835.21 
148.60 
13 
28.1 
11.88 
333.83 
789.61 
141.13 
14 
39.1 
9.57 
374.19 
1,528.81 
91.58 
15 
46.8 
10.94 
511.99 
2,190.24 
119.68 
16 
48.5 
9.58 
464.63 
2,352.25 
91.78 
17 
59.3 
10.09 
598.34 
3,516.49 
101.81 
18 
70 
8.11 
567.70 
4,900.00 
65.77 
19 
70 
6.83 
478.10 
4,900.00 
46.65 
20 
74.5 
8.88 
661.56 
5,550.25 
78.85 
21 
72.1 
7.68 
553.73 
5,198.41 
58.98 
22 
58.1 
8.47 
492.11 
3,375.61 
71.74 
23 
44.6 
8.86 
395.16 
1,989.16 
78.50 
24 
33.4 
10.36 
346.02 
1,115.56 
107.33 
25 
28.6 
11.08 
316.89 
817.96 
122.77 
Sum 
1315 
235.6 
11821.43 
76323.42 
2284.11 
From the above table, Σx = 1315, Σy = 235.6, Σxy = 11821.43, Σx^{2} = 76323.42, Σy^{2} = 2284.11 n is the sample size (25, in our case).
We will use the following formula for finding a and b, where the regression line is y = a + bx
a= 13.62, b= 0.079
Hence the regression line is y = 13.62  0.079x
The least square estimates for constant is 13.62 and slope is 0.079
Following are the residuals
Observation 
steam (x) 
pressure (y) 
xy 
x^2 
y^2 
estimated y 
Residual 
1 
35.3 
10.98 
387.59 
1,246.09 
120.56 
10.81 
0.17 
2 
29.7 
11.13 
330.56 
882.09 
123.88 
11.25 
0.12 
3 
30.8 
12.51 
385.31 
948.64 
156.50 
11.16 
1.35 
4 
58.8 
8.4 
493.92 
3,457.44 
70.56 
8.93 
0.53 
5 
61.4 
9.27 
569.18 
3,769.96 
85.93 
8.72 
0.55 
6 
71.3 
8.73 
622.45 
5,083.69 
76.21 
7.93 
0.80 
7 
74.4 
6.36 
473.18 
5,535.36 
40.45 
7.68 
1.32 
8 
76.7 
8.5 
651.95 
5,882.89 
72.25 
7.50 
1.00 
9 
70.7 
7.82 
552.87 
4,998.49 
61.15 
7.98 
0.16 
10 
57.5 
9.14 
525.55 
3,306.25 
83.54 
9.03 
0.11 
11 
46.4 
8.24 
382.34 
2,152.96 
67.90 
9.92 
1.68 
12 
28.9 
12.19 
352.29 
835.21 
148.60 
11.32 
0.87 
13 
28.1 
11.88 
333.83 
789.61 
141.13 
11.38 
0.50 
14 
39.1 
9.57 
374.19 
1,528.81 
91.58 
10.50 
0.93 
15 
46.8 
10.94 
511.99 
2,190.24 
119.68 
9.89 
1.05 
16 
48.5 
9.58 
464.63 
2,352.25 
91.78 
9.75 
0.17 
17 
59.3 
10.09 
598.34 
3,516.49 
101.81 
8.89 
1.20 
18 
70 
8.11 
567.70 
4,900.00 
65.77 
8.03 
0.08 
19 
70 
6.83 
478.10 
4,900.00 
46.65 
8.03 
1.20 
20 
74.5 
8.88 
661.56 
5,550.25 
78.85 
7.68 
1.20 
21 
72.1 
7.68 
553.73 
5,198.41 
58.98 
7.87 
0.19 
22 
58.1 
8.47 
492.11 
3,375.61 
71.74 
8.98 
0.51 
23 
44.6 
8.86 
395.16 
1,989.16 
78.50 
10.06 
1.20 
24 
33.4 
10.36 
346.02 
1,115.56 
107.33 
10.96 
0.60 
25 
28.6 
11.08 
316.89 
817.96 
122.77 
11.34 
0.26 
Sum 
1315 
235.6 
11821.432 
76323.42 
2284.1102 
235.6 
1.6E14 
For the ANOVA table following are needed to be calculated:
Observation 
steam (x) 
pressure (y) 
xy 
x^2 
y^2 
estimated y 
(yestiamted y)^2 
(estimated y avg. of y)^2 
(yavg. of y)^2 
Residual 
1 
35.3 
10.98 
387.59 
1,246.09 
120.56 
10.81 
0.03 
1.91 
2.42 
0.17 
2 
29.7 
11.13 
330.56 
882.09 
123.88 
11.25 
0.02 
3.35 
2.91 
0.12 
3 
30.8 
12.51 
385.31 
948.64 
156.50 
11.17 
1.81 
3.03 
9.52 
1.34 
4 
58.8 
8.4 
493.92 
3,457.44 
70.56 
8.93 
0.28 
0.24 
1.05 
0.53 
5 
61.4 
9.27 
569.18 
3,769.96 
85.93 
8.72 
0.30 
0.49 
0.02 
0.55 
6 
71.3 
8.73 
622.45 
5,083.69 
76.21 
7.93 
0.63 
2.22 
0.48 
0.80 
7 
74.4 
6.36 
473.18 
5,535.36 
40.45 
7.69 
1.76 
3.02 
9.39 
1.33 
8 
76.7 
8.5 
651.95 
5,882.89 
72.25 
7.50 
1.00 
3.69 
0.85 
1.00 
9 
70.7 
7.82 
552.87 
4,998.49 
61.15 
7.98 
0.03 
2.08 
2.57 
0.16 
10 
57.5 
9.14 
525.55 
3,306.25 
83.54 
9.03 
0.01 
0.15 
0.08 
0.11 
11 
46.4 
8.24 
382.34 
2,152.96 
67.90 
9.92 
2.82 
0.25 
1.40 
1.68 
12 
28.9 
12.19 
352.29 
835.21 
148.60 
11.32 
0.76 
3.58 
7.65 
0.87 
13 
28.1 
11.88 
333.83 
789.61 
141.13 
11.38 
0.25 
3.83 
6.03 
0.50 
14 
39.1 
9.57 
374.19 
1,528.81 
91.58 
10.50 
0.87 
1.16 
0.02 
0.93 
15 
46.8 
10.94 
511.99 
2,190.24 
119.68 
9.89 
1.11 
0.22 
2.30 
1.05 
16 
48.5 
9.58 
464.63 
2,352.25 
91.78 
9.75 
0.03 
0.11 
0.02 
0.17 
17 
59.3 
10.09 
598.34 
3,516.49 
101.81 
8.89 
1.44 
0.28 
0.44 
1.20 
18 
70 
8.11 
567.70 
4,900.00 
65.77 
8.04 
0.01 
1.92 
1.73 
0.07 
19 
70 
6.83 
478.10 
4,900.00 
46.65 
8.04 
1.46 
1.92 
6.73 
1.21 
20 
74.5 
8.88 
661.56 
5,550.25 
78.85 
7.68 
1.45 
3.05 
0.30 
1.20 
21 
72.1 
7.68 
553.73 
5,198.41 
58.98 
7.87 
0.04 
2.42 
3.04 
0.19 
22 
58.1 
8.47 
492.11 
3,375.61 
71.74 
8.99 
0.27 
0.19 
0.91 
0.52 
23 
44.6 
8.86 
395.16 
1,989.16 
78.50 
10.06 
1.45 
0.41 
0.32 
1.20 
24 
33.4 
10.36 
346.02 
1,115.56 
107.33 
10.96 
0.36 
2.35 
0.88 
0.60 
25 
28.6 
11.08 
316.89 
817.96 
122.77 
11.34 
0.07 
3.67 
2.74 
0.26 
Sum 
1315 
235.6 
11821.432 
76323.42 
2284.1102 
235.638 
18.2234617 
45.5596905 
63.8158 
0.038 
DF 
SS 
MS 
F 

Regression 
1 
SSR= ∑ ( 
MSR=SSR/1 
F^{∗}=MSR/MSE 
Residual 
n2 
SSE=∑(yi− 
MSE=SSE/(n−2) 

Total 
n1 
SSTO=∑(y_{i}− 

Where

df 
SS 
MS 
F 
Regression 
1 
45.5924 
45.5924 
57.54279 
Residual 
23 
18.2234 
0.792322 

Total 
24 
63.8158 


The ANOVA table us used for test hypotheses about the regression linearity or population means. When the null hypothesis of equal means is true, the two mean squares estimate the same quantity (error variance), and should be of approximately equal magnitude. In other words, their ratio should be close to 1.
Following is the formula for computing correlation coefficient
Replacing the value, r= 0.845
Coefficient of determination is r^{2} = 0.714
STD for
Predictor 
Error 
Slope 
Constant 
Constant 
62.95 
10.175 
6.142 
Waist 
0.874 
0.108 
8.123 
STD 
4.549 
8.86 
7.86 
For slope
We are testing the null hypothesis H_{0}: b = 0 against the alternative hypothesis Ha: b ≠ 0.
If b = 0 then F = 1
If F > 1 then b ≠ 0, which means there is a linear relationship
In view of the fact there F = 57.54, we can say the slope is significant
The confidence interval for Slope is (b1 is slope of the regression line, σb_{1 }is standard error estimate (SE))
Where
SE= sqrt [ Σ(yi  ?i)2 / (n  2) ] / sqrt [ Σ(xi  x)2 ]
Computing the value the CI is (0.109, 0.050)
Fo scatterplot of data following steps are followed in MInitab:
Choose Graph > Scatterplot.
Choose Simple, then click OK.
Under Y variables, select pressure as Y variable.
Under X variables, select steam as X variable.
Click OK.
There seems to be a negative correlation between the variables as can be seen from the scatterplot
For linear regression, following are the steps which are taken
Choose Stat > Regression > Fitted Line Plot.
In Response (Y), enter steam.
In Predictor (X), enter pressure.
Under Type of Regression Model, choose linear.
Click OK in each dialog box.
For calculating the residuals, following are the steps:
Select Stat >> Regression >> Fit Regression Model ...
Specify the response and the predictor(s).
Under Graphs...
Under Residuals for Plots, select Regular
Under Residuals Plots, select Residuals versus fits.
Select OK.
For ANOVA table, Stat > ANOVA > General Linear Model > Fit General Linear Model
For correlation coefficient, Click “Stat”, then click “Basic Statistics” and then click “Correlation.”
Coefficient of determination is r^{2} = 0.714
From the output of regression:
As we can see the P value is 0 for both of the constant and slope. This means that null hypothesis for both can be rejected which stated the constant = 0 and slope – 0
From the output of regression the CI can be identified
linearity and additivity – This can be seen from normal probability plot
statistical independence – This can be seen from the Versus fits
homoscedasticity  This can be seen from the Versus Orders
normality  This can be seen from the histogram