Question

In a certain type of metal test specimen, the normal stress on a specimen is known to be functionally related to shear resistance. The following is a set of coded experimental data on the two variables. Read the data provided in the table above, into R as two vectors. Produce an appropriate plot to observe the linear relationship between the two variables. Produce an appropriate linear regression model in R, for the data provided. Normal Stress, x Shear Resistance, y 26.8 26.5 25.4 27.3 28.9 24.2 23.6 27.1 27.7 23.6 23.9 25.9 24.7 26.3 28.1 22.5 26.9 21.7 27.4 21.4 22.6 25.8 25.6 24.9 a) Using an appropriate test statistic from your output, test the pair of hypotheses, HB = O versus H :B 0. You are required to carry out this using the critical value approach at a level of significance of a = 0.05 (4) b) Produce appropriate R code to find a 95% confidence interval for the regression slope coefficient. Paste your code and output in the space provided. Provide an interpretation of this interval.[3] c) Using values from your output, write out the estimated linear regression equation, is the estimated regression equation suitable for predicting the shear resistance for a normal stress of 24.5? Support your answer with appropriate summary statistics for the variable normal stress. If your answer is yes, find the predicted shear resistance for a normal stress of 24.5. Paste any R code and output used below. [6] d) Produce appropriate diagnostic plots in R to assess the normality and constant variance assumptions for the residuals of your regression model. Paste your code and output in the space provided. State your conclusions made from observing each plot in [6]

Answer 1

x=C(26.8,25.4,28.9,23.6,27.7,23.9,24.7,28.1,26.9,27.4,22.6,25. 6) the critical value for this test required is table value of t at 0.975 level of confi- dence with 10 df and it can be obtained as > qt (0.975, 10) [1] 2.228139 Since the test statistic is greater than critical value in magnitude), we reject the Ho and conclude that regression slope() is significant. y=c(26.5,27.3,24.2,27.1,23.6,25.9,26.3,22.5,21.7,21.4,25.8,24. 9) > model=lm (y^x) > summary (model) Call: lm (formula = y ~ x) Answer(b): The 95% confidence interval of B1 can be obtained as below: > confint (model, 'x', level=0.95) 2.5 % 97.5 % x -1.242908 -0.1292458 The 95% CI of B1 indicates that we are 95% confident that the B1 will lie be- tween -1.2429 and -0.1292. Also it does not include 0 which is the confirmation that it is significantly different from 0. Residuals: Min 1Q Median -2.42633 -0.92139 -0.04785 32 0.89367 Max 2.30506 Coefficients: Estimate Std. Error t value Pr>tl) (Intercept) 42.5818 6.5065 6.544 6.52e-05 *** -0.6861 0.2499 -2.745 0.0206 * - - - Signif. codes: 01***0.001 '**' 0.01 '*' 0.05 '.' 0.1'' 1 Residual standard error: 1.64 on 10 degrees of freedom Multiple R-squared: 0.4298, Adjusted R-squared: 0.3727 F-statistic: 7.537 on 1 and 10 DF, p-value: 0.02064 Answer(c): The estimated regression equation is y = 42.5818- 0.6861 * * The estimated model is suitable for predicting the shear resistance using normal stress value of 24.5. The summary of Normal Stress,x > summary (x) Min. 1st Qu. Median Mean 3rd Qu. Max. 22.60 24.50 26.20 25.97 27.48 28.90 From the above summary we can see that the independent variables in our model includes the values from 22.60 to 28.90. Hence our estimated model is suitable for predicting the shear resistance using normal stress value of 24.5 as it will be interpolation and our model have the trained to capture the pattern be- tween 22.60 and 28.90. Answer(a): We have to test HO: B1=0 Hl: B170 From the regression output of R we have test staitistic t=-2.745

> plot (model, which=1) The prediction of y when x=24.5 using R can be obtained as below > new=data.frame (cbind (x=24.5)) > predict (model, new) Residuals vs Fitted 25.77291 10 2 1 Answer(d): Normality Assumptions: for checking the normality assumption we can we can use the normal Q-Q plot. In R it is obtained as below. > plot (model, which=2) 0 Residuals / 0 -1 Normal Q-Q -2 1.5 -10 01090 1.0 -3 23 0.5 Standardized residuals 0.0 24 25 26 Fitted values Imly - x) In above plot we can see that there is no visible pattern of residuals along the mean line of residuals and also the scatterings of residuals seem to be constant which confirms the assumption of constant variance. -0.5 -1.0 -1.5 09. 010 End of document -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Theoretical Quantiles Im(y -X) From above plot we can see that most of the points are densely lying along the line which confirms the assumption of normality. Constant Variance: for checking the assumption of constant variance of residu- als we can use Residual vs. Fitted plot as below

> x=C (26.8,25.4,28.9,23.6,27.7,23.9,24.7,28.1,26.9,27.4,22.6,25.6) > y=c (26.5, 27.3, 24.2, 27.1,23.6,25.9,26.3, 22.5,21.7,21.4,25.8,24.9) > model=lm (y-x) > summary (model) Call: lm (formula = y - x) Residuals: Min 10 Median -2.42633 -0.92139 -0.04785 30 0.89367 30 2.30665 Max 2.30506 Coefficients: Estimate Std. Error t value Pr(>tl) (Intercept) 42.5818 6 .5065 6.544 6.52e-05 *** -0.6861 0.2499 -2.745 0.0206 * --- Signif. codes: 01***' 0.001 1**0.01 1*' 0.05 .' 0.1 il Residual standard error: 1.64 on 10 degrees of freedom Multiple R-squared: 0.4298, Adjusted R-sg F-statistic: 7.537 on 1 and 10 DF, p-value: 0.02064 VVX V > confint (model, 'x', level=0.95) 2.5 % 97.5 % x -1.242908 -0.1292458 > summary (x) Min. lst Qu. Median 22.60 24.50 26.20 Mean 3rd Qu. 25.97 27.48 Max. 28.90 V > new=data.frame (cbind (x=24.5)) > new 1 24.5 predict (model, new) V 25.77291 VN > plot (model, which=2) > plot (model, which=1)

R-code

x=c(26.8,25.4,28.9,23.6,27.7,23.9,24.7,28.1,26.9,27.4,22.6,25.6)
y=c(26.5,27.3,24.2,27.1,23.6,25.9,26.3,22.5,21.7,21.4,25.8,24.9)

model=lm(y~x)
summary(model)

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

summary(x)

new=data.frame(cbind(x=24.5))
new
predict(model,new)

plot(model,which=2)
plot(model,which=1)