Normal stress | 26 | 25 | 28 | 23 | 27 | 23 | 24 | 28 | 26 |

Shear Stress | 22 | 27 | 24 | 27 | 23 | 25 | 26 | 22 | 21 |

- Identify which one is response variable, and fit a simple regression line, assuming that the relationship between them is linear.
- Interpret the regression coefficient with reference to your problem.
- Obtain the coefficient of determination, and interpret this.
- Based on the fitted model in (a), predict the shear resistance for normal stress of 30 kilogram per square centimeter .

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Normal Stress(x) |
Shear(y) |
x^{2} |
y^{2} |
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 | Σx^{2} = 5908 |
Σy^{2} = 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 + bΣx^{2}

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 = γ^{2} = (-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.

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## Discussion