Question

"FATALS","CUTTING"

270,15692

183,16198

319,17235

103,18463

149,18959

124,19103

62,19618

298,20436

330,21229

486,18660

302,17551

373,17466

187,17388

347,15261

168,14731

234,14237

68,13216

162,12017

27,11845

40,11905

26,11881

41,11974

116,11892

84,11810

43,12076

292,12342

89,12608

148,13049

166,11656

32,13305

72,13390

27,13625

154,13865

44,14445

3,14424

3,14315

153,13761

11,12471

9,10960

17,9218

2,9054

5,9218

63,8817

41,7744

10,6907

3,6440

26,6021

52,5561

31,5309

3,5320

19,4784

10,4311

12,3663

88,3060

0,2779

41,2623

2,2058

5,1890

2,1535

0,1515

0,1595

23,1803

4,1495

0,1432

The above contains data on the following two variables

• FATALS: the annual number of fatalities from gas and dust explosions in coal mines for years 1915 to 1978.

• CUTTING: the number of cutting machines in use

(a) Fit the regression model using FATALS as the dependent variable and CUTTING as the independent variable.

(b) Using appropriate residual plots and formal tests, investigate the violation of any assumptions. Do any assumptions of the linear regression model appear to be violated? If so, which one (or ones)?

(Hint: Plot of residuals versus fitted values can be used for linearity, zero mean, and constant variance. Normal probability plot of the residuals can be used for normality. We also have formal tests for the constant variance and normality assumptions that you can do in R).

Hint: data=read.table(‘hmw6_prob3.txt’, header=T, sep=‘,’) y=data$FATALS x=data$CUTTING

Answer 1

As i just have the data with me i have manually entered the values of data in R and carried out the analysis the code and output is as follows

t=c(270,15692,183,16198,319,17235,103,18463,149,18959,124,19103,62,19618,298,20436,330,21229,486,18660,302,17551,373,17466,187,17388,347,15261,168,14731,234,14237,68,13216,162,12017,27,11845,40,11905,26,11881,41,11974,116,11892,84,11810,43,12076,292,12342,89,12608,148,13049,166,11656,32,13305,72,13390,27,13625,154,13865,44,14445,3,14424,3,14315,153,13761,11,12471,9,10960,17,9218,2,9054,5,9218,63,8817,41,7744,10,6907,3,6440,26,6021,52,5561,31,5309,3,5320,19,4784,10,4311,12,3663,88,3060,0,2779,41,2623,2,2058,5,1890,2,1535,0,1515,0,1595,23,1803,4,1495,0,1432) cutting=t[seq(2,128,2)] fatals=t[seq(1,127,2)] model=lm(fatals~cutting) model fatals_fitted=-47.70923+0.01343*cutting residual=fatals_fitted-fatals plot(residual,fatals_fitted) qqnorm(residual) shapiro.test(residual) 

The output is as follows:

model

Call:
lm(formula = fatals ~ cutting)

Coefficients:
(Intercept) cutting
-47.70923 0.01343

Hence model is

fatal=-47.70923+0.01343*cutting

As seen from this plot the variance is not constant hence assumption of constant variance is voilated.

to check whether the residuals have normal distribution i have used shapiro test further and qq plot as well

This is the required qq plot

The line in qq plot is not perfectly straight but can be considered straight to some extent and hence can be said that it has normal distribution.

Shapiro-Wilk normality test

data: residual
W = 0.95666, p-value = 0.02465

This is the Shapiro test from this it can be seen that p-value is 0.02465 so at 1%LOS we can say that residuals are normally distributed.