Question

USE R STUDIO

The stackloss data frame available in R contains 21 observations on four variables taken at a factory where ammonia is converted to nitric acid. The first three variables are Air.Flow, Water.Temp, and Acid.Conc. The fourth variable is stack.loss, which measures the amount of ammonia that escapes before being absorbed. Read the help file for more information about this data frame.

- Give a numerical summarization of each column of the dataset, then use boxplots to help illustrating the summarization.

- Use a pairwise scatterplot to explore possible relationships between acid concentration, water temperature, and air flow. Is there a linear relationship between any pair of variables? Write it as a comment in your R script file.

- Draw scatter plot the fourth variable versus each of the first three variables. Discover if there is any linear relationship between the acid concentration, the water temperature, the air flow and the amount of ammonia that escapes before being absorbed. Write it as a comment in your R script file.

- From the above part, you may discover multiple scatter plots indicating linear trends. Choose one of them (Air.Flow, Water.Temp, or Acid.Conc) as explanatory variable and build a simple linear regression model based on this variable and the response (stack.loss). You need to include at least the following in your code:

– Scatter plot with the linear regression line on the data.

– Write as comments the interpretation of the linear regression coefficients.

– Check for normality of residuals using QQ-plot. Write some comments on the normality check.

– Check constant variance of error using the residual plot. Write some comments on the variance of residuals.

– Check the goodness of fit by extracting and reporting the R2 value. Write some comments on the goodness of fit.

Answer 1

> data <- stackloss > # 1) Numerical summarization of each column of the dataset > summary (data) Air.Flow Min. Water.Temp Min. Acid.Conc. Min. stack.loss Min. :50.00 1st Qu.:56.00 Median :58.00 :17.0 :72.00 : 7.00 1st Qu. :11.00 Median :15.00 1st Qu.:18.0 Median :20.0 1st Qu.:82.00 Median :87.00 Mean :60.43 Mean :21.1 Mean :86.29 3rd Qu.:89.00 Mean :17.52 3rd Qu.:62.00 3rd Qu.:24.0 3rd Qu.:19.00 Max. :80.00 Max. :27.0 Max. :93.00 Max. :42.00 > # boxplot of each column of the dataset > boxplot (data) > # 2) pairwise scatterplot > plot (data) > # comment: in all the pairwise sctterplot there is moderately positive relationship present. > # 3) fourth vatiable versus all other variable2 scatter plot is the fourth row (or column) of pairwise scatter plot. > # qqplot for first and fourth column > qqplot (data SAir.Flow, data$stack.loss) > # data is normally distributed. > # qqplot for second and fourth column > qqplot (data$water.Temp, dataSstack.loss) > # data is normally distributed. > # qqplot for third and fourth column > qqplot (data$Acid.Conc.,data$stack.loss) > # data is not normally distributed.

> # 4) > # a) > plot (data$Water.Temp , data$stack.loss) > abline (1m (data$stack.loss~data$water.Temp)) > # there is a linear relationship between the water temperature and stack loss. > # b) > regression <- 1m(data$stack.loss > regression data$Air.Flow) Call: 1m(formula = data$stack.loss data$Air.Flow) Coefficients: (Intercept) -44.13 data$Air.Flow 1.02 > # intercept = -44.13 which means when there is no air flow then the stack loss is 44.13 > # slope = 1.02 which means if air floW will increase by 1 unit then stack flow will increase by 1.02 units. > # c) > qqplot (data$Water.Temp , data$stack.loss) > # data is a normally distributed. > # d) > residual <- resid(regression) > plot (data$SAir.Flow, residual) > abline (0,0) > # Residual plot shows a fiarly random pattern, hence variance of error is constant. > # e)

> # e) > summary(regression) Call: data$Air.Flow) 1m (formula = data$stack.loss Residuals: Min Median 3Q 1.1166 Max 1Q -1.1272 8.8728 |-0.0459 -12.2896 Coefficients: Estimate Std. Error t value Pr(> |t|) -7.228 7.31e-07 *** 10.208 3.77e-09 *** 6.10586 (Intercept) dataSAir. Flow -44.13202 0.09995 1.02031 0 *** 0.001 **' 0.01 *' 0.05 . ' 0.1 Signif. codes: Residual standard error: 4.098 on 19 degrees of freedom Adjusted R-s quared: p-value: 3.774e-09 Multiple R-s quared: F-statistic: 104.2 on 1 and 19 DF, 0.8458, 0.8377 > # R-Square = 0.8458, hence our regression model is good fit to the data
Note : just run the given rcode and plot the graphs.