Question

3.2. A Trent Honour’s Thesis student is doing a project examining the health of a native brook trout population in Cold Creek. He has collected length, weight and girth measurements on several hundred fish but only determined the fat content in 40 of the fish. He wants to use the girth and fat content data from the 40 fish to be able to predict the fat content of the other fish. He has asked you to help him construct this predictive model.

Girth (mm)	Fat Content (%)
90	7.2
90	8.5
87	7.8
85	7.5
84	8.9
81	8.7
76	7.6
75	6.2
75	8.2
70	6.9
70	7.7
69	7.6
67	8.6
62	5.6
61	5.3
60	6.7
59	7.1
56	8.0
53	6.1
52	4.4
51	5.2
50	6.5
50	7.1
48	3.8
46	5.5
43	7.8
41	6.0
37	6.8
36	5.1
36	3.9
33	2.6
31	5.4
29	4.1
29	4.8
27	2.2
22	4.7
22	6.4
21	2.7
20	2.9
20	3.0

A). Create a well labelled xy scatter plot on the spreadsheet page with the data; make note of the general pattern of the plotted data and whether there are any suspect data points.

B). Use the advanced regression analysis tool to conduct a complete simple linear regression analysis for the data. You will not require the options for "residuals" for this analysis. Remember to complete the five steps of hypothesis testing in the spaces provided on the worksheet.

C). Use the linear regression equation determined in part "B" to calculate a set of "predicted y values" for each observed x value.

D). Add this set of predicted y values (from part C) to the scatter plot created previously - be sure to show this new set of values as a "line" rather than as a set of "markers".

**please be sure to show all work and each of the steps for the hypothesis test so that I can learn and understand**

Answer 1

a)

....

b)

Regression Statistics
Multiple R	0.7782
R Square	0.6056
Adjusted R Square	0.5953
Standard Error	1.1936
Observations	40
ANOVA
	df	SS	MS	F	Significance F
Regression	1	83.14	83.14	58.36	0.0000
Residual	38	54.14	1.42
Total	39	137.28
	Coefficients	Standard Error	t Stat	P-value	lower 95%	upper 95%
Intercept	2.4787	0.501	4.943	0.0000	1.4636	3.49
X	0.067	0.009	7.639	0.0000	0.0494	0.0849

Ho:   β1=   0
H1:   β1╪   0
n=   40  
alpha =   0.05  
estimated std error of slope =Se(ß1) = Se/√Sxx =    1.1936/√18439.1=   0.0088
t stat = estimated slope/std error =ß1 /Se(ß1) =    (0.0671-0)/0.0088=   7.64
Degree of freedom ,df = n-2=   38  
      
p-value =    0.0000  
decison :    p-value<α , reject Ho  
Conclusion:   Reject Ho and conclude that slope is significantly different from zero  

..............

c

x	y				Ŷ
90	7.2				8.52
90	8.5				8.52
87	7.8				8.32
85	7.5				8.19
84	8.9				8.12
81	8.7				7.92
76	7.6				7.58
75	6.2				7.51
75	8.2				7.51
70	6.9				7.18
70	7.7				7.1791
69	7.6				7.1120
67	8.6				6.977657
62	5.6				6.641912
61	5.3				6.574763
60	6.7				6.507615
59	7.1				6.440466
56	8				6.239019
53	6.1				6.037572
52	4.4				5.970423
51	5.2				5.903275
50	6.5				5.836126
50	7.1				5.836126
48	3.8				5.701828
46	5.5				5.56753
43	7.8				5.366083
41	6				5.231786
37	6.8				4.96319
36	5.1				4.896041
36	3.9				4.896041
33	2.6				4.694595
31	5.4				4.560297
29	4.1				4.425999
29	4.8				4.425999
27	2.2				4.291701
22	4.7				3.955957
22	6.4				3.955957
21	2.7				3.888808
20	2.9				3.821659
20	3				3.821659

..................

Please let me know in case of any doubt.

Thanks in advance!

Please upvote!