Question

Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and workpiece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. An article gave the accompanying data on x = fluid-flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than 10 µm (mg/m³):

x	88	177	182	354	369	442	961
y	0.41	0.60	0.46	0.66	0.58	0.69	0.96

(b) What proportion of observed variation in mist can be attributed to the simple linear regression relationship between velocity and mist? (Round your answer to three decimal places.)

answer= 0.908

(c) 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.)

H₀: β₁ = 0.0006667
H_a: β₁ < 0.0006667

t=______________

P-value=_____________

(d) Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability. (Calculate a 95% CI. Round your answers to six decimal places.)( , ) mg/m³

Answer 1

X	Y	XY	X²	Y²
88	0.41	36.08	7744	0.1681
177	0.6	106.2	31329	0.36
182	0.46	83.72	33124	0.2116
354	0.66	233.64	125316	0.4356
369	0.58	214.02	136161	0.3364
442	0.69	304.98	195364	0.4761
961	0.96	922.56	923521	0.9216
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
2573	4.36	1901.2	1452559	2.9094

Sample size, n =	7
x̅ = Ʃx/n = 2573/7 =	367.5714286
y̅ = Ʃy/n = 4.36/7 =	0.622857143
SSxx = Ʃx² - (Ʃx)²/n = 1452559 - (2573)²/7 =	506797.7143
SSyy = Ʃy² - (Ʃy)²/n = 2.9094 - (4.36)²/7 =	0.193742857
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 1901.2 - (2573)(4.36)/7 =	298.5885714

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

= (298.58857)²/(506797.71429*0.19374) = 0.9080

c)

Null and alternative hypothesis:

Ho: β₁ = 0.0006667

Ha: β₁ < 0.0006667

n = 7

α = 0.05

Slope, b = SSxy/SSxx = 298.58857/506797.71429 = 0.000589167

Sum of Square error, SSE = SSyy -SSxy²/SSxx

= 0.19374 - (298.58857)²/506797.71429 = 0.017824276

Standard error, se = √(SSE/(n-2)) = √(0.01782/(7-2)) = 0.05971

Test statistic:

t = (b-β₁)/(se/√SSxx) = (0.000589 - 0.0006667)/(0.05971/√506797.7143) = -0.92

df = n-2 = 5

p-value = T.DIST.2T(ABS(-0.9244), 5) = 0.199

Conclusion:

p-value > α , Fail to reject the null hypothesis.

d) Critical value, t_c = T.INV.2T(0.05, 5) = 2.5706  

95% Confidence interval for slope:  

Lower limit = β₁ - tc*se/√SSxx = 0.0006 - 2.5706*0.0597/√506797.7143 = 0.000374

Upper limit = β₁ + tc*se/√SSxx = 0.0006 + 2.5706*0.0597/√506797.7143 = 0.000805