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.
b. What is the point estimate of the beta coefficient? (By R.) Interpret it.
c. What is s_e? (By R) Interpret it.
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.
e. Suppose the fluid velocity is 250 cm/sec. Find the mean of the corresponding y in a way that conveys information about precision and reliability. 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.

Step-by-step explanation:

a) image attached

b) Lets do the analysis in R , the complete R snippet is as follows

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

y<- c(.4,.6,.48,.66,.61,.69,.99)

# scatterplot

plot(x,y, col="red",pch=16)

# model

fit <- lm(y~x)

summary(fit)

#equation is

#y = 0.4041 + 0.0006211*X

# beta coeffiecients are

fit$coefficients

coef(summary(fit))[, "Std. Error"]

# confidence interval of slope

confint(fit, 'x', level=0.95)

The results are

> summary(fit)

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 ***

---

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 # model is able to capture 93% of the variation of the data

F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403 , p value is less than 0.05 , hence model as a whole is significant

> fit$coefficients

(Intercept) x

0.4041237853 0.0006210758

> coef(summary(fit))[, "Std. Error"]

(Intercept) x

3.458905e-02 7.579156e-05

> confint(fit, 'x', level=0.95)

2.5 % 97.5 %

x 0.0004262474 0.0008159042

c)

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

> y=c(0.40,0.60,0.48,0.66,0.61,0.69,0.99)

>

> ### linear model

> model=lm(y~x)

> summary(model)

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 ***

---

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

s_e is the Residual standard error from the model and its estimated value is 0.05405. s_e is the standard deviation of the model.  

d) 95% confidence interval

> confint(model, confidence=0.95)

2.5 % 97.5 %

(Intercept) 0.3152097913 0.4930377793

x 0.0004262474 0.0008159042

Comment: The estimated confidence interval of slope of x does not include zero. Hence, x has the significant effect on y at 0.05 level of significance.

 e)

> predict(model, newdata=data.frame(x=250), interval="confidence", level=0.95)

fit lwr upr

1 0.5593927 0.5020485 0.616737

f)

> predict.lm(model, newdata=data.frame(x=250), interval="prediction", level=0.95)

fit lwr upr

1 0.5593927 0.4090954 0.7096901

the anwers is yes.

step-by-step explanation:

for part a. the answer is 5 miles in a hour. in part b. the answer is 10 miles in a hour. sandra runs a mile in 6 minutes while tall runs a mile in 12 minutes