In a certain type of metal test specimen. the effect of normal stress on a specimen is known to be functionally related to shear resistance. The following table gives the data on the two variables.

Normal stress 26 25 28 23 27 23 24 28 26
Shear Stress 22 27 24 27 23 25 26 22 21
  1. Identify which one is response variable, and fit a simple regression line, assuming that the relationship between them is linear.
  2. Interpret the regression coefficient with reference to your problem.
  3. Obtain the coefficient of determination, and interpret this.
  4. Based on the fitted model in (a), predict the shear resistance for normal stress of 30 kilogram per square centimeter .

This answer is restricted. Please login to view the answer of this question.

Login Now

Solution:

Normal Stress(x) Shear(y) x2 y2 xy
26 22 676 484 572
25 27 625 729 675
28 24 784 576 672
23 27 529 729 621
27 23 729 529 621
23 25 529 625 575
24 26 576 676 624
28 22 784 484 616
26 21 676 441 546
Σx = 230 Σy = 217 Σx2 = 5908 Σy2 = 5273 Σxy = 5522

n = 9

Solution i)

We already assumed that the relationship between Normal stress(x) and shear resistance (y) is linear.

The response variable (dependent) is shear resistance(y).

Now,

The regression Equation of Y on x is

y = a + bx —– (i)

The normal equation of Equatin(i) are

Σy = na + bΣx

217 = 9a + 230b   ——- (ii)

Σxy = a Σx + bΣx2

5522 = 230a + 5908b   ——– (iii)

Solving Eqn(ii) and (iii), We get

b = -0.85

and a = 45.84

Therefore, The fitted regrssion line is

y = a + bx

y = 45.84 + (-0.85)x

Solution ii)

Here, the regresion cofficient are

a = 45.84, means it is the value of shear resistance(y) when Normal stress (x) is 0. b = -0.85 means shear resistance 9y) changes by 0.85 per unit change in Normal Stress (x).

Solution iii)

First, We know

Correlation cofficient is given by

= \(\frac{n\sum xy – \sum x \sum y}{\sqrt{n\sum x^2 – (\sum x)^2} . \sqrt{n\sum y^2 – (\sum y)^2}}\)

= \(\frac{9 \times 5522 – 230 \times 217}{\sqrt{9 \times 5908 – (230)^2} . \sqrt{9 \times 5273 – (217)^2}}\)

= -0.67

Therefore, Cofficient determination = γ2 = (-0.67)2 = 0.4489

It means 44.89% of variation on shear resistance (γ) is explained by normal stress(x).

Solution iv)

Here, we know, Y = 45.84 – 0.85x

From Question,

x = 30

y = 45.84 – 0.85 x 30

= 20.34 ≈ 20

Therefore, The shear resistance for normal stress of 30kg per square is 20.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .