Question

1. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.)

Karlsruhe Method	Lehigh Method
1.2150	1.1419
1.2418	1.1595
1.2218	1.1546
1.2539	1.1821
1.2179	1.2219
1.1907	1.2204
1.2472	1.1607
1.1921	1.1775

Draw the normal probability plots for two samples using Microsoft Excel Data Analysis Toolpak. What are their R-squared?

Answer 1

Ho: there is no significant difference in the average strength between Karlsruhe and Lehigh procedures. U1 = u2

H1: there is a significant difference in the average strength between Karlsruhe and Lehigh procedures. U1 = u2

	Sample 1	Sample 2
n=	8.000	8.000	COUNT
mean=	1.223	1.177	AVERAGE
s=	0.02385	0.02983	STDEV
s^2/n	0.0001	0.0001

Sp^2
{(n1-1)*s1^2 + (n2-1)*s2^2}/{n1+n2-2}
=((8-1)*0.02385^2+(8-1)*0.02983^2)/(8+8-2)
0.0007

T=
(Xbar1-Xbar2)/sqrt(Sp^2*(1/n1+1/n2))
(1.22255-1.17733)/SQRT(0.0001*(1/8+1/8))
3.3489

Df= n1+n2-2
8+8-2
14

P-value
2*(1-P(T<|t|)
2*(1-P(T<abs(3.3489))
T.DIST.2T(abs(3.3489),14)
0.0052

With (t=3.34, p<5%),the null hypothesis is rejected at 5% level of significance. Hence there is sufficient evidence to conclude that there is a significant difference in the average strength between Karlsruhe and Lehigh procedures. U1 = u2

The normal probability plot is given below:

The value of coefficient of determination, R square is 14.6%

The regression output is given below: