Question

you can see the answer a, b, and c. I have a question about d, e, and f only how to solve theses questions

Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and work-piece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article "Variables Affecting Mist Generation from Metal Removal Fluids" (Lubrication Engr., 2002: 10-17) 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 some value:

x: 89 177 189 354 362 442 965

y: .40 .60 .48 .66 .61 .69 .99

a. Make a scatterplot of the data. By R.

x = c(89, 177, 189, 354, 362, 442, 965)

y = c(0.4, 0.6, 0.48, 0.66, 0.61, 0.69, 0.99)

plot(x,y)

b. What is the point estimate of the beta coefficient? (By R.) Interpret it.

lm.1 = lm(y~x)

lm.1

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x

0.4041238    0.0006211

0.4041238 =   

0.0006211 =

c. What is s_e? (By R) Interpret it.

summary(lm.1)

Call:

lm(formula = y ~ x)

Residuals:

      1 2 3        4 5 6        7

-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346

Coefficients:

            Estimate Std. Error t value Pr(>|t|)    

(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***

x           6.211e-04 7.579e-05   8.195 0.00044 ***

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.05405 on 5 degrees of freedom

Multiple R-squared: 0.9307,   Adjusted R-squared: 0.9168

F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403

Residual standard error: 0.05405

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. Hint: This question is asking for a CI for beta. Compute it AND interpret it. By hand; i.e. you must use the basic formulas for the CI. E.g. for beta: beta_hat +- t* s_e/sqrt(S_xx) , but you may use R to compute the various terms in the formula. Use 95% confidence level.

e. Suppose the fluid velocity is 250 cm/sec. Compute an interval estimate of the corresponding mean y value. Use 95% confidence level. Interpret the resulting interval. By hand, as in part d.

f. Suppose the fluid velocity for a specific fluid is 250 cm/sec. Predict the y for that specific fluid in a way that conveys information about precision and reliability. Use 95% prediction level. Interpret the resulting interval. By hand, as in part d.

Answer 1

D ) Here we find confidence interval for Slope : b1

Formula :   

where b1 = coefficient of b1 = 0.0006211 , Sb1 = standard error of b1 = 7.579e-05 = 0.0000757 ,

t_n-2 is the critical value at t distribution with df = n-2 = 7-2=5 and c-level = 0.95 .

So here critical value tn-2 = 2.571

Lower limit :

Upper limit :

the true average change in mist associated with a 1 cm/sec increase in velocity is lies between (0.000426 to 0.000816) .

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e. Suppose the fluid velocity is 250 cm/sec. Compute an interval estimate of the corresponding mean y value. Use 95% confidence level. Interpret the resulting interval.

Fluid velocity = x= 250 .

Formula :

where

S_xx= 508479.4 and we know t_n-2 = 2.571

Use regression line to find y^ ,y^ = 0.404124+0.000612*x

                             ===== >>> y^ = 0.404124 + ( 0.000621* 250) =0.5594

Lower limit =

Upper limit =

For the fluid velocity is 250 cm/sec the 95% confidence interval estimate of the corresponding mean y value is (-0.4116 , 0.1504 )

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f. Suppose the fluid velocity for a specific fluid is 250 cm/sec. Predict the y for that specific fluid in a way that conveys information about precision and reliability. Use 95% prediction level. Interpret the resulting interval.

Formula :

Lower limit =

Upper limit =

For the fluid velocity is 250 cm/sec the 95% he y for that specific fluid in a way that conveys information about precision and reliability is lies(0.4109,0.7079)

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