Question

5. You have been given the task of determining if the number of board feet of lumber (Volume) from a black cherry tree can be predicted from the diameter of the tree at 54 inches from the ground. You have data from 31 black cherry trees in New Zealand about their volume, height, and diameter. You need to analyze the data and write a couple of sentences discussing the questions below. Follow the steps we use in class to analyze the simple regression model. (13 points) Try to run the following charts, scatter plots, statistics, etc. to support the use of the linear regression model for this situation. (3 points) a. Scatterplot, IV vs. DV with regression line and r2 (Excel) (1 point) b. Residuals (Excel), Scatterplot, Residuals vs. IV (residual plot) (Excel) (1 point) c. Regression Statistics (b0 and b1, multiple r, r2, SST, SSR, and SSE) highlighted in your regression output (1 point) Analyze the model and answer the following questions in paragraph form. Give me a few sentences as justification using the values from your output. (10 points) a. Can you predict volume with diameter using a linear regression model? Were any violations of the assumptions of linear regression found? (2 points) b. How accurate is your model? (i.e., What is the coefficient of determination?) (2 points) c. What is the magnitude of the relationship between volume and diameter? (2 points) d. What would you predict volume to be if diameter is 17.4 inches? (2 points) e. Is there a statistically significant relationship between last volume and diameter? (2 points )

Diam	Height	Volume
8.3	70	10.3
8.6	65	10.3
8.8	63	10.2
10.5	72	16.4
10.7	81	18.8
10.8	83	19.7
11.0	66	15.6
11.0	75	18.2
11.1	80	22.6
11.2	75	19.9
11.3	79	24.2
11.4	76	21.0
11.4	76	21.4
11.7	69	21.3
12.0	75	19.1
12.9	74	22.2
12.9	85	33.8
13.3	86	27.4
13.7	71	25.7
13.8	64	24.9
14.0	78	34.5
14.2	80	31.7
14.5	74	36.3
16.0	72	38.3
16.3	77	42.6
17.3	81	55.4
17.5	82	55.7
17.9	80	58.3
18.0	80	51.5
18.0	80	51.0
20.6	87	77.0

Answer 1

Solution:

a. Can you predict volume with diameter using a linear regression model? Were any violations of the assumptions of linear regression found?

Yes, we can predict the volume with diameter using a linear regression model, because the p-value for this regression model is given as 0.00 which is less than 5% level of significance or alpha value 0.05. There are no any violations of the assumptions of linear regression found. The regression coefficients for this regression models are statistically significant as the p-values for these coefficients are less than significance level 0.05.

b. How accurate is your model? (i.e., what is the coefficient of determination?

The value for the R square or coefficient of determination is given as 0.9353, which means about 93.53% of the variation in the dependent variable volume is explained by the independent variable diameter. So, we can say that given regression model is 93.53% accurate.

c. What is the magnitude of the relationship between volume and diameter?

The correlation coefficient between the dependent variable volume and independent variable diameter is given as 0.9671, this means there is a strong positive linear association or relationship or correlation exists between the given two variables volume and diameter.

d. What would you predict volume to be if diameter is 17.4 inches?

From given regression output, we are given a regression model for prediction of volume as below:

Volume = -36.9435 + 5.0659*Diameter

We are given Diameter = 17.4

Volume = -36.9435 + 5.0659*17.4

Volume = 51.20316

e. Is there a statistically significant relationship between last volume and diameter?

There is a statistically significant relationship exists between the two variables volume and diameter, because the p-value for the regression model is given as 0.00 approximately. This p-value is less than alpha value 0.05 and it is indicated that there is a statistically significant relationship exists between two variables.

Excel Outputs and Explanation:

Here, we have to analyze given data for three variables diameter, height, and volume. We have to use descriptive statistics, scatter plots, correlation coefficients, regression analysis, residual plots, etc. We have to construct the regression model for the prediction of dependent variable or response variable volume based on the independent variable diameter. Also, we have to see the regression model for the prediction of response variable volume based on the independent variables diameter and height. We have to use Excel data analysis for analyzing this data. Excel outputs for this analysis is given as below:

Descriptive statistics for the given three variables are summarised as below:

	Diameter	Height	Volume
Mean	13.2483871	76	30.17096774
Standard Error	0.563626334	1.144411386	2.952324375
Median	12.9	76	24.2
Mode	11	80	10.3
Standard Deviation	3.138138617	6.371812929	16.43784644
Sample Variance	9.847913978	40.6	270.2027957
Kurtosis	-0.435421384	-0.45110847	0.772668312
Skewness	0.553465175	-0.39420581	1.119260015
Range	12.3	24	66.8
Minimum	8.3	63	10.2
Maximum	20.6	87	77
Sum	410.7	2356	935.3
Count	31	31	31

Scatter plots between different variables are given as below:

Correlation coefficients between the different variables are summarised as below:

	Diameter	Height	Volume
Diameter	1
Height	0.51928007	1
Volume	0.96711937	0.59824965	1

Regression model for the prediction of dependent variable Volume based on the independent variable diameter is given as below:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.967119368
R Square	0.935319872
Adjusted R Square	0.933089523
Standard Error	4.251987522
Observations	31
ANOVA
	df	SS	MS	F	Significance F
Regression	1	7581.781332	7581.781332	419.3602785	8.64433E-19
Residual	29	524.3025387	18.07939788
Total	30	8106.083871
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-36.94345912	3.365144948	-10.97826682	7.62145E-12	-43.82595322	-30.06096503
Diameter	5.065856423	0.24737695	20.47828798	8.64433E-19	4.559913759	5.571799086
RESIDUAL OUTPUT
Observation	Predicted Volume	Residuals
1	5.103149185	5.196850815
2	6.622906112	3.677093888
3	7.636077396	2.563922604
4	16.24803332	0.151966685
5	17.2612046	1.5387954
6	17.76779024	1.932209758
7	18.78096153	-3.180961527
8	18.78096153	-0.580961527
9	19.28754717	3.312452831
10	19.79413281	0.105867189
11	20.30071845	3.899281546
12	20.8073041	0.192695904
13	20.8073041	0.592695904
14	22.32706102	-1.027061023
15	23.84681795	-4.74681795
16	28.40608873	-6.20608873
17	28.40608873	5.39391127
18	30.4324313	-3.032431299
19	32.45877387	-6.758773868
20	32.96535951	-8.065359511
21	33.9785308	0.521469205
22	34.99170208	-3.29170208
23	36.51145901	-0.211459007
24	44.11024364	-5.810243641
25	45.63000057	-3.030000568
26	50.69585699	4.704143009
27	51.70902828	3.990971725
28	53.73537084	4.564629156
29	54.24195649	-2.741956487
30	54.24195649	-3.241956487
31	67.41318319	9.586816814

Residual plot for this regression model is given as below:

Regression model for the prediction of dependent variable Volume based on the independent variables diameter and height is given as below:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.973627258
R Square	0.947950038
Adjusted R Square	0.944232183
Standard Error	3.881832038
Observations	31
ANOVA
	df	SS	MS	F	Significance F
Regression	2	7684.162512	3842.081256	254.9723374	1.07124E-18
Residual	28	421.9213592	15.06861997
Total	30	8106.083871
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-57.98765892	8.638225865	-6.712913024	2.74951E-07	-75.68226224	-40.2930556
Diameter	4.708160503	0.264264609	17.81608409	8.2233E-17	4.166838997	5.249482009
Height	0.339251234	0.130151181	2.606593597	0.014490974	0.07264863	0.605853839
RESIDUAL OUTPUT
Observation	Predicted Volume	Residuals
1	4.837659654	5.462340346
2	4.553851633	5.746148367
3	4.816981266	5.383018734
4	15.87411523	0.525884771
5	19.86900844	-1.069008438
6	21.01832696	-1.318326957
7	16.19268807	-0.592688075
8	19.24594918	-1.045949183
9	21.4130214	1.186978595
10	20.18758128	-0.287581284
11	22.01540227	2.184597729
12	21.46846462	-0.468464619
13	21.46846462	-0.068464619
14	20.50615413	0.79384587
15	23.95410969	-4.854109686
16	27.8522029	-5.652202905
17	31.58396648	2.216033519
18	33.80648192	-6.406481917
19	30.6009776	-4.900977604
20	28.69703501	-3.797035015
21	34.38818439	0.111815605
22	36.00831896	-4.308318964
23	35.38525971	0.914740291
24	41.768998	-3.468997996
25	44.87770232	-2.277702318
26	50.94286776	4.457132242
27	52.22375109	3.476248908
28	53.42851283	4.871487175
29	53.89932888	-2.399328876
30	53.89932888	-2.899328876
31	68.51530482	8.484695177