MATH1312 Regression Analysis Proof Reading Service

MATH1312 Regression Analysis Assignment Help

MATH1312 Regression Analysis Proof Reading Service

1 A.

Scatter plot:

MATH1312 Regression Analysis Assignment Help

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.

1 B.

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	x²	y²
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, Σx² = 356828, Σy² = 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

MATH1312 Regression Analysis Assignment Help

a= 6.13, b= 0.006

Hence the regression line is y = 6.127 + 0.005x

MATH1312 Regression Analysis Assignment Help

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1C.

For the ANOVA test, following table is needed to be calculated:

	DF	SS	MS	F
Regression	1	SSR= ∑ (yi−y)²	MSR=SSR/1	F^∗=MSR/MSE
Residual	n-2	SSE=∑(yi−yi)²	MSE=SSE/(n−2)
Total	n-1	SSTO=∑(y_i−y)2

Whereyiis the estimated valued through the regression line

Following is the calculation for the table

Observation	Area (x)	ph (y)	xy	x^2	y^2	estimated y	(y-estimated y)^2	(estimated y- avg. of y)^2	(y-avg. 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 H₀: 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

1D.

MATH1312 Regression Analysis Assignment Help

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

1E.

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:

MATH1312 Regression Analysis Assignment Help

Computing the values, following is the confidence interval (6.88, 29.47)

2 A.

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, Σx² = 76323.42, Σy² = 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

MATH1312 Regression Analysis Assignment Help

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

2B.

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	-1.6E-14

2C.

For the ANOVA table following are needed to be calculated:

Observation	steam (x)	pressure (y)	xy	x^2	y^2	estimated y	(y-estiamted y)^2	(estimated y- avg. of y)^2	(y-avg. 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= ∑ (yi−y)²	MSR=SSR/1	F^∗=MSR/MSE
Residual	n-2	SSE=∑(yi−yi)²	MSE=SSE/(n−2)
Total	n-1	SSTO=∑(y_i−y)2

Whereyiis the estimated valued through the regression line

	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.

2D.

Following is the formula for computing correlation coefficient

MATH1312 Regression Analysis Assignment Help

Replacing the value, r= -0.845

Coefficient of determination is r² = 0.714

2E.

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

2F.

For slope

We are testing the null hypothesis H₀: 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

2G.

The confidence interval for Slope is (b1 is slope of the regression line, σb₁is standard error estimate (SE))

MATH1312 Regression Analysis Assignment Help

Where

SE= sqrt [ Σ(yi - ?i)2 / (n - 2) ] / sqrt [ Σ(xi - x)2 ]

Computing the value the CI is (-0.109, -0.050)

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3A.

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.

MATH1312 Regression Analysis Assignment Help

There seems to be a negative correlation between the variables as can be seen from the scatterplot

3 B.

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.

MATH1312 Regression Analysis Assignment Help

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.

MATH1312 Regression Analysis Assignment Help

For ANOVA table, Stat > ANOVA > General Linear Model > Fit General Linear Model

MATH1312 Regression Analysis Assignment Help

For correlation coefficient, Click “Stat”, then click “Basic Statistics” and then click “Correlation.”

MATH1312 Regression Analysis Assignment Help

Coefficient of determination is r² = 0.714

From the output of regression:

MATH1312 Regression Analysis Assignment Help

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

MATH1312 Regression Analysis Assignment Help