THE SIMPLE LINEAR REGRESSION

SIMPLE LINEAR REGRESSION (SLR) AND ANALYSIS OF VARIANCE (ANOVA)
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SIMPLE LINEAR REGRESSION (SLR) AND ANALYSIS OF VARIANCE (ANOVA)
Question 1
Using the Young Modulus dataset and each of the parallel (E_11) and perpendicular (E_22)
datasets, the following were the results of the Simple Linear Regression analysis where fiber
volume fraction was the explanatory variable.
a. The regression line parameters were:
Table 1: Regression Linear Parameters for both Datasets.
Coefficients E_22
Intercept 3.964386
Fiber Volume
Fraction 0.084211
b. The coefficients of determination (𝑅2) were 0.9965 for E_11 and 0.4359 for E_22.
Table 2: Test statistics calculated for both datasets to determine the level of evidence.
c. The decision of whether or not you reject the null hypothesis that the slope was zero,
and the level of evidence against the null (numerical and verbal conclusion needed).
Coefficients E_11
Intercept 1.888175
Fiber Volume
Fraction 0.323768
Regression Statistics E_11
Multiple R 0.998248
R Square 0.9965
Adjusted R Square 0.995625
Standard Error 0.330613
Observations 6
Regression Statistics E_22
Multiple R 0.660201
R Square 0.435865
Adjusted R Square 0.294831
Standard Error 1.650703
Observations 6

SIMPLE LINEAR REGRESSION (SLR) AND ANALYSIS OF VARIANCE (ANOVA)
Test of hypothesis:
H0: B0=0, that is, the coefficients were equal to 0.
H1: B0 ≠ 0, otherwise.
Table 3:The P-values for both datasets.
E_11 Coefficients P-value
Intercept 1.888175 0.011726
Fiber
Volume
Fraction 0.323768 4.6E-06
E_22
Coefficient
s P-value
Intercept 3.964386
0.13820
4
Fiber Volume
Fraction 0.084211
0.15357
8
At p-values of 0.011726 and 4.6e^06, the slope B0 was not equal to zero and the two
coefficients were significantly different from 0. Also, they were statistically significant in
predicting E_11.
For E_22 prediction, the regression’s p-values were greater than the 0.05 significance
level. The null hypothesis was rejected. The slope, B0, was different from 0. Therefore,
the coefficients were considered insignificantly different from 0 and insignificant in
predicting E_22.
d. Provide a 95% Confidence Interval for the mean response for a Fiber Volume
Fraction of 35%, and a Prediction Interval for a future observation to be taken with
a Fiber Volume Fraction of 35%.
Using Excel and basic statistical concepts of deriving confidence intervals, the basic
formula is usually:

SIMPLE LINEAR REGRESSION (SLR) AND ANALYSIS OF VARIANCE (ANOVA)
Lower interval = Mean of Response(Y) – t-critical value*standard error
Upper interval = Mean of Response(Y) + t-critical value*standard error
Where the standard error is:
S.E = Standard deviation / sqrt (n)
Where n is the sample size.
The E_11 simple linear regression equation was:
E_11 = 1.888175 + 0.323768 Fiber Volume Fraction
For a Fiber Volume Fraction of 35%, the response value would be:
E_11 = 1.888175 + 0.323768*35
E_11= 1.888175 + 11.33188
E_11= 13.220055
The following were the 95% confidence intervals for mean of E_11: C.I. = (10.77942,
19.82345).
For E_22, the linear regression equation was:
E_22 = 3.964386 + 0.084211Fibre Volume Fraction
For a Fiber Volume of 35%, the E_22 value would be:
E_22 = 3.964386 + 0.084211*35
E_22 = 3.964386 + 2.947385
E_22 = 6.911771
Similar to previous procedure of obtaining confidence intervals, the 95% confidence
intervals were: C.I. = (5.694283, 9.211937). See the table below for more calculations.
Table 4: Confidence Intervals for Both Datasets.
Fiber Volume FractionE_11 E_22 CI for mean E_11 and E_22 for Fibre Vol. 35%
20 8.49 5.75 n 7 7
30 11.7 7.11 df 5 5
40 14.3 7.18 mean(y) 15.30143714 7.45311
50 18.2 5.42 x0 35 35
55 20 10.19 y0 13.220055 6.91177
60 21.2 9.61 std 4.654249598 1.81026
35 13.2201 6.91177 se 1.759140997 0.68421
t-crit 2.570581836 2.57058
lower 10.77942125 5.69428
upper 19.82345304 9.21194