6-46. The data shown in Table 6-15 represent the thrust of a jet-turbine engine (y) and six candidate regressors: x₁ = pri- mary speed of rotation, x₂ = secondary speed of rotation, x3 fuel flow rate, x₁ pressure, x, = exhaust temperature, and x = ambient temperature at time of test. = = X6 (a) Fit a multiple linear regression model using x, = fuel flow rate, X4 = pressure, and x = exhaust temperature as the regressors. (b) Test for significance of regression using a = 0.01. Find

The dataset contains observationsDependent variable:: Thrust of jet, turbine engine.Independent variables (Regressors):: Primary speed of rotation.: Secondary speed of rotation.: Fuel flow rate.: Pressure.: Exhaust temperature.: Ambient temperature.The R-code for fitting of the multiple linear regression model:1# Given data 2 3y=c(4540,4315,4095,3650,3200,4833,4617,4340,3820,3368,4445,4188,3981,3622,3125,4560,4340,4114,3630,3210,4330,4119,3891,3467,3045,4411,4203,3968,3531,3074,4350,4128,3940,3480,3064,4402,4180,3973,3530,3080) 4 5X3=c(30250,30010,29780,29330,28960,30083,29831,29604,29088,28675,30120,29920,29720,29370,28940,30160,29960,29710,29250,28890,30190,29960,29770,29360,28960,30160,29940,29760,29350,28910,30180,29940,29750,29360,28900,30170,29950,29740,29320,28910) 6 7X4=c(205,195,184,164,144,215,206,195,171,149,195,190,180,161,139,208,199,187,164,145,193,183,173,153,134,193,184,173,153,133,198,186,178,156,136,197,188,178,156,137) 8 9X5=c(1732,1697,1662,1598,1541,1709,1669,1640,1572,1522,1740,1711,1682,1630,1572,1704,1679,1642,1576,1528,1748,1713,1684,1624,1569,1746,1714,1679,1621,1561,1729,1692,1667,1609,1552,1758,1729,1690,1616,1569) 10 11# (a) Fit a multiple linear regression model using X3, X4, X5 as regressors. 12 13model=lm(y ~ X3 + X4 + X5) 14summary(model)The output of the code:> summary(model)Call:lm(formula = y ~ X3 + X4 + X5)Residuals: Min 1Q Median 3Q Max -77.340 -26.631 -5.083 14.481 85.014 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4210.6811 2327.5427 1.809 0.07880 . X3 -0.2313 0.1119 -2.068 0.04591 * X4 21.4846 0.9381 22.903 < 2e-16 ***X5 1.6916 0.5633 3.003 0.00484 ** ---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 44.63 on 36 degrees of freedomMultiple R-squared: 0.9928, Adjusted R-squared: 0.9922 F-statistic: 1650 on 3 and 36 DF, p-value: < 2.2e-16Therefore the multiple linear regression equation for the model using as regressors is,.The R-code for the test for significance of regression using .1# (b) Test for significance of regression using alpha=0.01. 2 3# Find the p-value for this test. 4 5anova_result=anova(model) 6 7p_value=anova_result$`Pr(>F)`[1] 8 9# Display the p-value 10 11print(p_value) 12 13# Check if the p-value is less than alpha=0.01 for significance 14 15if (p_value < 0.01) { 16 17cat("The regression model is significant.\n") 18 19} else { 20 21cat("The regression model is not significant.\n") 22 23} 24 25The output of above code-> print(p_value)[1] 4.618411e-39> # Check if the p-value is less than alpha=0.01 for significance> if (p_value < 0.01) {+ cat("The regression model is significant.\n")+ } else {+ cat("The regression model is not significant.\n")+ }The regression model is significant.The p-value of the test is therefore the regression model is significant.The R-code for finding the t-test statistics for each regressor1# (c) Find the t-test statistic for each regressor. 2 3summary_result=summary(model) 4 5# Extract t-test statistics and p-values 6 7t_stats = summary_result$coefficients[, "t value"] 8 9p_values=summary_result$coefficients[, "Pr(>|t|)"] 10 11# Display t-test statistics and p-values 12 13print(data.frame(Regressor = rownames(summary_result$coefficients), t_stats, p_values)) 14 15# Check significance using alpha=0.01 16 17significant_regressors=names(t_stats[p_values < 0.01]) 18 19cat("Significant regressors:", paste(significant_regressors, collapse = ", "), "\n")The output of the above Code-> print(data.frame(Regressor = rownames(summary_result$coefficients), t_stats, p_values)) Regressor(Intercept) (Intercept)X3 X3X4 X4X5 X5 t_stats(Intercept) 1.809067X3 -2.067706X4 22.902840X5 3.003081 p_values(Intercept) 7.879793e-02X3 4.591245e-02X4 4.709029e-23X5 4.837918e-03> cat("Significant regressors:", paste(significant_regressors, collapse = ", "), "\n")Significant regressors: X4, X5 The pressure and Exhaust temperature are the significant regressors.The R-code for the and the adjusted :1# (d) Find R-square and adjusted R-square for this model. 2 3rsquare=summary_result$r.squared 4 5adjusted_rsquare=summary_result$adj.r.squared 6 7# Display R-square and adjusted R-square 8 9cat("R-square:", rsquare, "\n") 10 11cat("Adjusted R-square:", adjusted_rsquare, "\n")The output of the above code-> cat("R-square:", rsquare, "\n")R-square: 0.9927812> cat("Adjusted R-square:", adjusted_rsquare, "\n")Adjusted R-square: 0.9921796Interpretation of both these values:The value of is and the value of adjusted is .The of means that approximately 99.28% of the variability in the thrust of jet turbine engine can be explained by the variation in fuel flow rate, pressure, and exhaust temperature included in the model.Adjusted is modified version of R Square that accounts for the number of predictors in the model. It penalized the model for including irrelevant predictors that do not significantly improve the fit. An adjusted of means that, after adjusting for the number of predictors, approximately 99.22% of the variability in dependent variable is still explained by the model.The R-code for the construction of the Normal probability plot of the model,1# (e) Construct a normal probability plot of the residuals and interpret this graph. 2residuals = resid(model) 3 4# Set up a new graphics window with a specified size 5options(repr.plot.width=4, repr.plot.height=4) # Adjust the width and height as needed 6 7# Create the normal probability plot 8qqnorm(residuals) 9qqline(residuals) 10 11# Reset the graphics options to default 12options(repr.plot.width=NULL, repr.plot.height=NULL)The normal probability plot of the residual:Interpretation of the graph-The points fall approximately on a straight line therefore the residuals are normally distributed.The R-code for the plot of the residuals versus 1# (f) Plot the residuals versus yhat. Are there any indications of inequality of variances or nonlinearity? 2yhat=predict(model) 3plot(yhat, residuals) 4abline(h = 0, col = "red", lty = 2)The output of the R-code-The plot of residuals versus Interpretation of the plot of the residual versus fitted value plot.Since the spread of residencies is roughly constant across predicted values, the assumption of equal variance is met.The R-code for the plot of residual versus :1# (g) Plot the residuals versus X3. Is there any indication of nonlinearity? 2plot(X3, residuals) 3abline(h = 0, col = "red", lty = 2) # Interpretation: If there's a clear pattern in the residuals against X3, there might be nonlinearity.The output of the above R-code:The plot of the residuals versus Interpretation:Since there is no clear pattern in the plot the data is linear.The R-code for predicting the thrust for the given values of the regressor variable:1# (h) Predict the thrust on engine for which X3=20000, X4=170, X5=1589 2new_data = data.frame(X3 = 20000, X4 = 170, X5 = 1589) 3prediction = predict(model, newdata = new_data) 4 5# Display the predicted thrust 6cat("Predicted Thrust:", prediction, "\n")The output of the above code:> cat("Predicted Thrust:", prediction, "\n")Predicted Thrust: 5924.175 The thrust for an engine for which is .(a) Therefore the multiple linear regression equation for the model using as regressors is,.(b) The p-value of the test is therefore the regression model is significant.(c) The pressure and Exhaust temperature are the significant regressors.(d) The value of is and the value of adjusted is .The of means that approximately 99.28% of the variability in the thrust of jet turbine engines can be explained by the variation in fuel flow rate, pressure, and exhaust temperature included in the model.Adjusted is modified version of R Square that accounts for the number of predictors in the model. It penalized the model for including irrelevant predictors that do not significantly improve the fit. An adjusted of means that, after adjusting for the number of predictors, approximately 99.22% of the variability in dependent variable is still explained by the model.(e) The normal probability plot of the residual:The points fall approximately on a straight line therefore the residuals are normally distributed.(f) The plot of residual versus :Since the spread of residencies is roughly constant across predicted values, the assumption of equal variance is met.(g) Plot of the residual versus :Since there is no clear pattern in the plot the data is linear.(h) The thrust for an engine for which is .