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	90	177	192	354	365	442	960
y	0.38	0.60	0.49	0.66	0.63	0.69	0.91

(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.)

c) find t value and 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
90	0.38	34.2	8100	0.1444
177	0.60	106.2	31329	0.36
192	0.49	94.08	36864	0.2401
354	0.66	233.64	125316	0.4356
365	0.63	229.95	133225	0.3969
442	0.69	304.98	195364	0.4761
960	0.91	873.6	921600	0.8281

Ʃx =	2580
Ʃy =	4.36
Ʃxy =	1876.65
Ʃx² =	1451798
Ʃy² =	2.8812
Sample size, n =	7
x̅ = Ʃx/n = 2580/7 =	368.5714286
y̅ = Ʃy/n = 4.36/7 =	0.622857143
SSxx = Ʃx² - (Ʃx)²/n = 1451798 - (2580)²/7 =	500883.7143
SSyy = Ʃy² - (Ʃy)²/n = 2.8812 - (4.36)²/7 =	0.165542857
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 1876.65 - (2580)(4.36)/7 =	269.6785714

b)

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

= (269.67857)²/(500883.71429*0.16554) = 0.8771

c)

Slope, b = SSxy/SSxx = 269.67857/500883.71429 = 0.0005384

y-intercept, a = y̅ -b* x̅ = 0.62286 - (0.00054)*368.57143 = 0.4244162

Regression equation :

ŷ = 0.4244 + (0.0005384) x

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 0.16554 - (269.67857)²/500883.71429 = 0.020346418

Standard error, se = √(SSE/(n-2)) = √(0.02035/(7-2)) = 0.0638

Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

Test statistic:

t = b/(se/√SSxx) = 5.9734

df = n-2 = 5

p-value = T.DIST.2T(ABS(5.9734), 5) = 0.0019

Conclusion:

p-value < α, Reject the null hypothesis.

d) true average change in mist associated with a 1 cm/sec increase in velocity = 0.000538

--

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

95% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = 0.0005 - 2.5706*0.0638/√500883.7143 = 0.000307

Upper limit = β₁ + tc*se/√SSxx = 0.0005 + 2.5706*0.0638/√500883.7143 = 0.000770