Question

A publisher of books has produced seven comparable Statistical Management books with the following costs.

Quantity produced (000)	1	2	4	5	7	9	13
Manufacturing Cost(000£)	5	5.9	6.5	7.5	8	9.5	10.8

a) (5 pts) Construct the regression line for predicting manufacturing costs from quality produced.

b) (5 pts) Calculate the predicted values and the residuals of this data.

c) (5 pts) Construct a 99% two-sided confidence interval for the intercept.

d) (5 pts) Construct a 99% two-sided confidence interval for the slope.

e) (5 pts) Test for significance of regression by ANOVA and interpret your conclusion.

Answer 1

No	x	y	(x-xbar)^2	(y-ybar)^2	(x-xbar)*(y-ybar)
1	1	5	23.591837	6.76	12.62857143
2	2	5.9	14.877551	2.89	6.557142857
3	4	6.5	3.4489796	1.21	2.042857143
4	5	7.5	0.7346939	0.01	0.085714286
5	7	8	1.3061224	0.16	0.457142857
6	9	9.5	9.877551	3.61	5.971428571
7	13	10.8	51.020408	10.24	22.85714286
sum	41	53.2	104.85714	24.88	50.6
mean	5.857142857	7.6	sxx	syy	sxy

calculation below using excell

slope = b1 = sxy/sxx = 0.482561308
intercept = b0 = ybar-(slope*xbar) = 4.773569482
SST = SYY = 24.88
SSR = sxy^2/sxx = 24.41760218
SSE = syy-sxy^2/sxx = 0.46239782
r^2 = SSR/SST = 0.981414879
r = sxy/sqrt(sxx*syy) = 0.990663858
error variance s^2 = SSE/(n-2) = 0.092479564
S^2b1 = s^2/sxx = 0.000881958
standard error b1 = se(b1) = sqrt(s^2b1) = 0.029697773
test statistics = b1/se(b1) = 16.24907413
For (1-alpha)% CI value of t = 0.99 = 4.032142984
Lower Confidence bound for slope = estimated slope -t*se(b1) = 0.362815643
Upper Confidence bound for slope = estimated slope +t*se(b1) = 0.602306973
standard error b0 = se(b0) = s*sqrt(1/n+ Xbar^2/Sxx) = 0.208489604
Lower Confidence bound for Intercept = estimated intercept -t*se(b0) = 3.932909588
Lower Confidence bound for Intercept = estimated intercept +t*se(b0) = 5.614229376

a) (5 pts) Construct the regression line for predicting manufacturing costs from quality produced.

slope = b1 = sxy/sxx = 0.482561308 =
intercept = b0 = ybar-(slope*xbar) = 4.773569482 =

y= 4.7736 + 0.4826*x

y=Manufacturing Cost

x=Quantity produced

Regression Model is

Manufacturing Cost =4.7736 + 0.4826*Quantity produced

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b)

Observation	x	y	Predicted y	Residuals
1	1	5	5.256131	-0.25613
2	2	5.9	5.738692	0.161308
3	4	6.5	6.703815	-0.20381
4	5	7.5	7.186376	0.313624
5	7	8	8.151499	-0.1515
6	9	9.5	9.116621	0.383379
7	13	10.8	11.04687	-0.24687

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c) 99% two-sided confidence interval for the intercept.

intercept = b0 = ybar-(slope*xbar) = 4.7736

For (1-alpha)% CI value of t (1-alpha=0.99) = 4.0321

standard error b0 = se(b0) = s*sqrt(1/n+ Xbar^2/Sxx) = 0.208489604
Lower Confidence bound for Intercept = estimated intercept -t*se(b0) = 3.9329
Lower Confidence bound for Intercept = estimated intercept +t*se(b0) = 5.6142

99% two-sided confidence interval for the Intercept=(3.9329 ,5.6142)

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d)99% two-sided confidence interval for the slope

slope = b1 = 0.4826

standard error b1 = se(b1) = sqrt(s^2b1) = 0.029697773
For (1-alpha)% CI value of t (1-alpha=0.99) = 4.0321
Lower Confidence bound for slope = estimated slope -t*se(b1) = 0.3628
Upper Confidence bound for slope = estimated slope +t*se(b1) = 0.6023

99% two-sided confidence interval for the slope =(0.3628 , 0.6023)

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e)

Test for significance of regression by ANOVA and interpret your conclusion.

ANOVA
	df	SS	MS	F	Significance F
Regression	1	24.41760218	24.4176	264.0324	1.61E-05
Residual	5	0.46239782	0.09248
Total	6	24.88

Null and alternative Hypothesis

H0: Model is not significant

H1:Overall model is significant

test statistics= F0= MS_Regression/MS_Residual =264.0324

p-value = p(F > F0)=0

Decision:P-value is less close to 0 hence we reject H0

Conclusion : Overall Model is significant

ANOVA
	df	SS	MS	F	Significance F
Regression	1	24.41760218	24.4176	264.0324	1.61E-05
Residual	5	0.46239782	0.09248
Total	6	24.88

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