Mist (airborne droplets or aerosols) is generated when metal-removing fluids are
ID: 3316543 • Letter: M
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):
(a) The investigators performed a simple linear regression analysis to relate the two variables. Does a scatter plot of the data support this strategy?
Yes, a scatter plot shows a reasonable linear relationship. [Correct]
No, a scatter plot does not show a reasonable linear relationship.
(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) The investigators were particularly interested in the impact on mist of increasing velocity from 100 to 1000 (a factor of 10 corresponding to the difference between the smallest and largest x values in the sample). When x increases in this way, is there substantial evidence that the true average increase in y is less than 0.6? (Use = 0.05.)
State the appropriate null and alternative hypotheses.
H0: 1 = 0.0006667
Ha: 1 > 0.0006667H0: 1 0.0006667
Ha: 1 = 0.0006667 H0: 1 = 0.0006667
Ha: 1 < 0.0006667H0: 1 = 0.0006667
Ha: 1 0.0006667 [Correct]
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.)
(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/m^3
x 87 177 185 354 366 442 963 y 0.39 0.60 0.46 0.66 0.61 0.69 0.93Explanation / Answer
a.
Yes, a scatter plot shows a reasonable linear relationship
b.proportion of observed variation in mist can be attributed to the simple linear regression relationship between velocity and mist
calculation procedure for regression
mean of X = X / n = 367.7143
mean of Y = Y / n = 0.62
(Xi - Mean)^2 = 508631.428
(Yi - Mean)^2 = 0.18
(Xi-Mean)*(Yi-Mean) = 286.819
b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= 286.819 / 508631.428
= 0.001
bo = Y / n - b1 * X / n
bo = 0.62 - 0.001*367.7143 = 0.413
value of regression equation is, Y = bo + b1 X
Y'=0.413+0.001* X
c.
Given that,
mean(x)=367.7143
standard deviation , s.d1=291.1562
number(n1)=7
y(mean)=0.62
standard deviation, s.d2 =0.174
number(n2)=7
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =367.7143-0.62/sqrt((84771.9328/7)+(0.03028/7))
to =3.336
| to | =3.336
critical value
the value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.447
we got |to| = 3.3358 & | t | = 2.447
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 3.3358 ) = 0.016
hence value of p0.05 > 0.016,here we reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
H0: 1 = 0.0006667
Ha: 1 0.0006667
test statistic: 3.336
critical value: -2.447 , 2.447
decision: reject Ho
p-value: 0.016
we have enough evidence to support the claim
d.
TRADITIONAL METHOD
given that,
mean(x)=367.7143
standard deviation , s.d1=291.1562
number(n1)=7
y(mean)=0.62
standard deviation, s.d2 =0.174
number(n2)=7
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((84771.932798/7)+(0.030276/7))
= 110.046719
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 6 d.f is 2.446912
margin of error = 2.447 * 110.046719
= 269.284322
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (367.7143-0.62) ± 269.284322 ]
= [97.809978 , 636.378622]
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DIRECT METHOD
given that,
mean(x)=367.7143
standard deviation , s.d1=291.1562
sample size, n1=7
y(mean)=0.62
standard deviation, s.d2 =0.174
sample size,n2 =7
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 367.7143-0.62) ± t a/2 * sqrt((84771.932798/7)+(0.030276/7)]
= [ (367.0943) ± t a/2 * 110.046719]
= [97.809978 , 636.378622] mg/m^3
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interpretations:
1. we are 95% sure that the interval [97.809978 , 636.378622] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
Line of Regression Y on X i.e Y = bo + b1 X X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean) 87 0.39 78800.518 0.053 64.564 177 0.6 36371.944 0 3.814 185 0.46 33384.515 0.026 29.234 354 0.66 188.082 0.002 -0.549 366 0.61 2.939 0 0.017 442 0.69 5518.365 0.005 5.2 963 0.93 354365.065 0.096 184.539