Homework SourseThe data given provides shear strengths for Ibeams Column A
The data given provides shear strengths for I-beams. Column A is from the I-beam currently used, column B is from a proposed substitute. Prove if the shear strength for the proposed substitute is greater than that currently used (90% confidence level).
A B
| 2110.46 | 2003.898 |
| 2083.236 | 2161.299 |
| 1894.69 | 2070.339 |
| 1899.14 | 2078.5 |
| 1979.78 | 2050.213 |
| 2020.605 | 2236.334 |
| 1919.594 | 1987.827 |
| 1939.767 | 1895.548 |
| 2008.881 | 2098.094 |
| 1965.564 | 2024.152 |
| 2039.726 | 2024.725 |
| 2007.004 | 2048.784 |
| 1899.521 | 2069.437 |
| 2090.526 | 1916.153 |
| 2008.903 | 2048.208 |
| 1709.704 | 1984.075 |
| 1989.777 | |
| 2074.655 | |
| 1784.298 | |
| 1899.8 |
Solution
Set Up Hypothesis
Null, shear strength for the proposed substitute is similar Ho: u1 = u2
Alternate, shear strength for the proposed substitute is greater than current - H1: u1 < u2
Test Statistic
X(Mean)=1969.78
Standard Deviation(s.d1)=101.66 ; Number(n1)=20
Y(Mean)=2043.59
Standard Deviation(s.d2)=83.181; Number(n2)=16
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1969.78-2043.59/Sqrt((10334.7556/20)+(6919.07876/16))
to =-2.4
| to | =2.4
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 15 d.f is 1.753
We got |to| = 2.39575 & | t | = 1.753
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value:Left Tail - Ha : ( P < -2.3957 ) = 0.01504
Hence Value of P0.05 > 0.01504,Here we Reject Ho
