ACC6025 Accounting Theory, Research and Practice 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