Question

The following data is representative of that reported in an article with x = burner-area liberation rate (MBtu/hr-ft²) and y = NO_x emission rate (ppm):

x	100	125	125	150	150	200	200	250	250	300	300	350	400	400
y	150	150	180	210	180	310	280	410	440	430	400	610	610	680

(a) Does the simple linear regression model specify a useful relationship between the two rates? Use the appropriate test procedure to obtain information about the P-value, and then reach a conclusion at significance level 0.01.
State the appropriate null and alternative hypotheses.

H₀: β₁ ≠ 0
H_a: β₁ = 0H₀: β₁ = 0
H_a: β₁ ≠ 0    H₀: β₁ = 0
H_a: β₁ < 0H₀: β₁ = 0
H_a: β₁ > 0

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t	=
P-value	=

State the conclusion in the problem context.

Fail to reject H₀. There is evidence that the model is useful.Reject H₀. There is no evidence that the model is useful.    Reject H₀. There is evidence that the model is useful.Fail to reject H₀. There is no evidence that the model is useful.

Answer 1

X	Y	XY	X²	Y²
100	150	15000	10000	22500
125	150	18750	15625	22500
125	180	22500	15625	32400
150	210	31500	22500	44100
150	180	27000	22500	32400
200	310	62000	40000	96100
200	280	56000	40000	78400
250	410	102500	62500	168100
250	440	110000	62500	193600
300	430	129000	90000	184900
300	400	120000	90000	160000
350	610	213500	122500	372100
400	610	244000	160000	372100
400	680	272000	160000	462400

Ʃx =	3300
Ʃy =	5040
Ʃxy =	1423750
Ʃx² =	913750
Ʃy² =	2241600
Sample size, n =	14
x̅ = Ʃx/n = 3300/14 =	235.7142857
y̅ = Ʃy/n = 5040/14 =	360
SSxx = Ʃx² - (Ʃx)²/n = 913750 - (3300)²/14 =	135892.8571
SSyy = Ʃy² - (Ʃy)²/n = 2241600 - (5040)²/14 =	427200
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 1423750 - (3300)(5040)/14 =	235750

Slope, b = SSxy/SSxx = 235750/135892.85714 =    1.734822602

y-intercept, a = y̅ -b* x̅ = 360 - (1.73482)*235.71429 =    -48.92247043
  
Sum of Square error, SSE = SSyy -SSxy²/SSxx = 427200 - (235750)²/135892.85714 = 18215.5716

Standard error, se = √(SSE/(n-2)) = √(18215.57162/(14-2)) =    38.96106

Null and alternative hypothesis:  

Ho: β₁ = 0  

Ha: β₁ ≠ 0  

n =    14

Test statistic:  

t = b/(se/√SSxx) =    16.41

df = n-2 =    12

p-value = T.DIST.2T(ABS(16.4143), 12) =    0.000

Conclusion:  

Reject H0. There is evidence that the model is useful.