Question

The data below contains the total costs and depths of 16 offshore oil wells. It is expected that cost is a linear function of the depth.

(a) Write out the equation of the regression line. Interpret the slope and intercept in the context of this problem. Do they make sense?

b) Test the hypothesis that the slope parameter is zero 4 different ways (ANOVA, t-test for β1, t-test for ρ, and a confidence interval for β1).

(c) What is the R2 for the SLR you have obtained? What does it mean?

(d) Plot the standardized residuals against the independent variable. What can you say about the regression using this graph? (HINT: Are there outliers? Does it seem reasonable to claim the data has a linear fit?)

Depth(feet) vs Cost($1000)

Depth (feet)	Cost ($1000)
5000	2596.800049
5200	3328
6000	3181.100098
6538	3198.399902
7109	4779.899902
7556	5905.600098
8005	5769.200195
8207	8089.5
8210	4813.100098
8600	5618.700195
9026	7736
9197	6788.299805
9926	7840.799805
10813	8882.5
13800	10489.5
14311	12506.59961

Answer 1

Depth (feet), X	Cost ($1000), Y	XY	X²	Y²
5000	2596.800049	12984000.25	25000000	6743370.494
5200	3328	17305600	27040000	11075584
6000	3181.100098	19086600.59	36000000	10119397.83
6538	3198.399902	20911138.56	42745444	10229761.93
7109	4779.899902	33980308.4	50537881	22847443.07
7556	5905.600098	44622714.34	57093136	34876112.52
8005	5769.200195	46182447.56	64080025	33283670.89
8207	8089.5	66390526.5	67354849	65440010.25
8210	4813.100098	39515551.8	67404100	23165932.55
8600	5618.700195	48320821.68	73960000	31569791.88
9026	7736	69825136	81468676	59845696
9197	6788.299805	62431993.31	84584809	46081014.24
9926	7840.799805	77827778.86	98525476	61478141.58
10813	8882.5	96046472.5	116920969	78898806.25
13800	10489.5	144755100	190440000	110029610.3
14311	12506.59961	178981947	204804721	156415033.8
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
137498	101523.9998	979168137.4	1287960086	762099377.6

Sample size, n =	16
x̅ = Ʃx/n = 137498/16 =	8593.625
y̅ = Ʃy/n = 101523.9998/16 =	6345.249985
SSxx = Ʃx² - (Ʃx)²/n = 1287960086 - (137498)²/16 =	106353835.8
SSyy = Ʃy² - (Ʃy)²/n = 762099377.55589 - (101523.99976)²/16 =	117904219.6
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 979168137.36836 - (137498)(101523.99976)/16 =	106708955

a)

Slope, b = SSxy/SSxx = 106708954.95661/106353835.75 = 1.003339035

y-intercept, a = y̅ -b* x̅ = 6345.24998 - (1.00334)*8593.625 = -2277.069432

Regression equation :

ŷ = -2277.0694 + (1.0033) x

Slope interpretation: A unit increase in depth will increase the cost by 1.0033 units

Y -Intercept: this the value at x = 0. this value is not reasonable because at x = 0, the cost is negative which cannot be true.

b)

Anova test:

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

SSE = SSyy - SSxy²/SSxx =10838959.72

SSR = SSxy²/SSxx = 107065259.9188

Test statistic:

F = SSR/(SSE/(n-2)) = 107065259.9188/(10838959.7209/14) = 138.2894

P-value = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.

Slope Hypothesis test:

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

Slope, b = 1.003339035

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 117904219.63968 - (106708954.95661)²/106353835.75 = 10838959.72

Standard error, se = √(SSE/(n-2)) = √(10838959.7209/(16-2)) = 879.89284

Test statistic:

t = b/(se/√SSxx) = 11.7597

df = n-2 = 14

p-value = T.DIST.2T(ABS(11.7597), 14) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.

Correlation Hypothesis test:

Null and alternative hypothesis:

Ho: ρ = 0

Ha: ρ ≠ 0

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 106708954.95661/√(106353835.75*117904219.63968) = 0.9529

Test statistic :  

t = r*√(n-2)/√(1-r²) = 0.9529 *√(16 - 2)/√(1 - 0.9529²) = 11.7597

df = n-2 = 14

p-value = T.DIST.2T(ABS(11.7597), 14) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis. There is a correlation between x and y.

95% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = 0.8203

Upper limit = β₁ + tc*se/√SSxx = 1.1863

As the confidence interval do not contain 0, we reject the null hypothesis.

c)

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy)

= (106708954.95661)²/(106353835.75*117904219.63968) = 0.9081

90.81% variation in y is due to the linear relationship between y and x variables.

d) Residuals:

X	Y	Predicted value, ŷ	Residual, y-ŷ
5000	2596.8	-2277.0694 + (1.0033) * 5000 = 2739.6257	-142.8257
5200	3328	-2277.0694 + (1.0033) * 5200 = 2940.2936	387.7064
6000	3181.1	-2277.0694 + (1.0033) * 6000 = 3742.9648	-561.8647
6538	3198.4	-2277.0694 + (1.0033) * 6538 = 4282.7612	-1084.3613
7109	4779.9	-2277.0694 + (1.0033) * 7109 = 4855.6678	-75.7679
7556	5905.6	-2277.0694 + (1.0033) * 7556 = 5304.1603	601.4398
8005	5769.2	-2277.0694 + (1.0033) * 8005 = 5754.6595	14.5406
8207	8089.5	-2277.0694 + (1.0033) * 8207 = 5957.334	2132.1660
8210	4813.1	-2277.0694 + (1.0033) * 8210 = 5960.344	-1147.2439
8600	5618.7	-2277.0694 + (1.0033) * 8600 = 6351.6463	-732.9461
9026	7736	-2277.0694 + (1.0033) * 9026 = 6779.0687	956.9313
9197	6788.3	-2277.0694 + (1.0033) * 9197 = 6950.6397	-162.3399
9926	7840.8	-2277.0694 + (1.0033) * 9926 = 7682.0738	158.7260
10813	8882.5	-2277.0694 + (1.0033) * 10813 = 8572.0356	310.4644
13800	10489.5	-2277.0694 + (1.0033) * 13800 = 11569.0093	-1079.5093
14311	12506.6	-2277.0694 + (1.0033) * 14311 = 12081.7155	424.8841