question archive In a certain type of metal test specimen, the normal stress on a specimen is known to be functionally related to shear resistance
In a certain type of metal test specimen, the normal stress on a specimen is known to be functionally related to shear resistance
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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 Ras 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 values from your output, write out the estimated linear regression equation.
b) Using an appropriate test statistic from your output, test the pair of hypotheses, Ho: B = 0 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
c) 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.
d) Is the estimated regression equation in (a) 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.
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Answer:
R codes and output:
> 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)
> fit=lm(y~x)
> summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-2.42633 -0.92139 -0.04785 0.89367 2.30506
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 42.5818 6.5065 6.544 6.52e-05 ***
x -0.6861 0.2499 -2.745 0.0206 *
---
Signif. codes: 0 ‘***’ 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
> qt(0.975,10) # critical value
[1] 2.228139
> confint(fit)
2.5 % 97.5 %
59(Intercept) 28.084338 57.0792671
x -1.242908 -0.1292458
> par(mfrow=c(2,2))
> plot(fit)
Que.a
Estimated linear regression equation is,
Y = 42.5818 - 0.6861 X
Que.b
Test statistic, t = -2.745
Critical value = 2.228139
Since calculated |t| = 2.745 is greater than critical value, hence we reject null hypothesis and conclude that slope coefficient is statistically significant.
Que.c
95% confidence interval for slope coefficient is ( -1.242908 , -0.1292458 ).
We are 95% confident that slope coefficient will lie within the interval ( -1.242908 , -0.1292458 ).
Que.d
> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
22.60 24.50 26.20 25.97 27.48 28.90
Since X=24.5 is less than maximum value , we can predict shear resistance for this value.
Y = 42.5818 - 0.6861 X
Y = 42.5818 - 0.6861 (24.50)
Y = 25.7724
