Question

1. SIMPLE LINEAR REGRESSION. For this and the next 3 parts. The journal, Fisheries Science (Feb 1995) reported on a study of the variables that affect endogenous nitrogen excretion (ENE) in carp raised in Japan. Carp were divided into groups of 2 to 15 fish each according to body weight and each group placed in a separate tank. The carp were then fed a protein-free diet three times daily for a period of 20 days. One day after terminating the feeding experiment, the amount of ENE in each tank was measured. The table below gives the mean body weight (in grams) and ENE amount (in milligrams per 100 grams of body weight per day) for each carp group [Source: Watanabe, T. and Ohta, M. "Endogenous Nitrogen Excretion and Non-Fecal Energy Loss in Carp and Rainbow Trout," Fisheries Science, Vol. 61, No. 1, Feb 1995, p. 56]. You should be able to determine which variable is the response variable and which is the explanatory variable. Which of the following is true? [I] Plot of the residuals shows an upward curvature [II] Plot of the residuals shows a downward curvature [III] The regression is significant at the 1% level [IV] Correlation coefficient of the two variables = -0.68 [V] Standard error of the estimate is 0.0297.

   Tank	Body Weight	ENE
   1	11.7	15.3
   2	25.3	9.3
   3	90.2	6.5
   4	213.0	6.0
   5	10.2	15.7
   6	17.6	10.0
   7	32.6	8.6
   8	81.3	6.4
   9	141.5	5.6
   10	285.7	6.0

   		I and V only
   		II, IV, V
   		I and IV only
   		II and III only
   		None of the above

Part B.

1. SIMPLE LINEAR REGRESSION (above data). Give a 99% prediction interval for the expected (mean) value of the dependent variable with X = 0 (note alpha = 0.01).

   		8.3724; 14.4355
   		6.9928; 15.8151
   		-0.0508; -0.0034
   		-0.0615; 0.0073
   		None of the above

Part C.

1. SIMPLE LINEAR REGRESSION (above data). Give a 99% confidence interval for the slope (note alpha = 0.01).

   		8.3724; 14.4355
   		6.9928; 15.8151
   		-0.0508; -0.0034
   		-0.0615; 0.0073
   		None of the above

Part D.

1. MULTIPLE REGRESSION (refer to above data). Conduct a multiple regression by introducing a quadratic term to the model. Which of the following is true? [I] About 74% of the variation in Y explained the regression [II] At the 1% level, the regression is statistically significant [III] At the 5% level, the regression is statistically significant [IV] Both coefficients are statistically significant at the 5% level [V] At least one of the independent variables is not significant

   		I, III, V
   		III, IV, V
   		I, II, IV
   		None of the above

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
11.7	15.3	6274.2241	40.4496	-503.776
25.3	9.3	4304.6721	0.1296	-23.6196
90.2	6.5	0.5041	5.9536	1.7324
213	6	14905.968	8.6436	-358.945
10.2	15.7	6514.1041	45.6976	-545.6
17.6	10	5374.3561	1.1236	-77.7086
32.6	8.6	3400.0561	0.1156	19.8254
81.3	6.4	92.3521	6.4516	24.4094
141.5	5.6	2559.3481	11.1556	-168.971
285.7	6	37943.144	8.6436	-572.683

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	909.1	89.4	81368.73	128.364	-2205.33
mean	90.91	8.94	SSxx	SSyy	SSxy

SSE=   (Sx*Sy - S²xy)/Sx =    68.59
      
std error ,Se =    √(SSE/(n-2)) =    2.9282
      
correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.6824
SSR=   S²xy/Sxx =   59.77

Anova table
variation	SS	df	MS	F-stat	p-value
regression	59.77	1	59.77	6.9711	0.0297
error,	68.59	8	8.57
total	128.36	9

part A)

answer : I and IV only

part B)

X Value=   0
Confidence Level=   99%
  
Intermediate Calculations  
Sample Size , n=   10
Degrees of Freedom,df=n-2 =   8
critical t Value=tα/2 =   3.355
X̅ =    90.910
Σ(x-x̅)² =Sxx   81368.729
Standard Error of the Estimate,Se=   2.928
h Statistic = (1/n+(X-X̅)²/Sxx) =    0.202
Predicted Y (YHat)   11.404
standard error, S(ŷ)=Se*√h stat =    1.3146424

pediction interval
margin of error,E=t*std error=t/S(ŷ)=    10.7699
Prediction Interval Lower Limit=Ŷ -E =   0.6340
Prediction Interval Upper Limit=Ŷ +E =   22.1738

so,answer: none of the above

part c)

confidence interval for slope      

n=   10              
alpha=   0.01              
estimated std error of slope =Se(ß1) =                s/√Sxx =    0.0103
slope ,    ß1 = SSxy/SSxx =   -0.027102967
t critical value=   t α/2 =    3.355387331      
margin of error ,E=       t*std error =        0.034443597
lower confidence limit =        ß̂1-E =   -0.061546564  
upper confidence limit=       ß̂1+E =   0.00734063  
answer: (-0.0615; 0.0073)

part D)

excel output

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.858733
R Square	0.737423
Adjusted R Square	0.6624
Standard Error	2.194327
Observations	10
ANOVA
	df	SS	MS	F	Significance F
Regression	2	94.65852	47.32926	9.829403	0.009277
Residual	7	33.70548	4.815069
Total	9	128.364
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	13.71274	1.306249	10.49779	1.55E-05	10.62395	16.80153	10.62395	16.80153
Body Weight	-0.10184	0.028811	-3.53474	0.009537	-0.16997	-0.03371	-0.16997	-0.03371
X^2	0.000273	0.000102	2.69174	0.031008	3.32E-05	0.000514	3.32E-05	0.000514

answer:I, II, IV