Question

A research reports a comparison of 2 methods for predicting the strength of 9 different types of steel. The following table shows the results of this research. a) We wish to determine whether there is any difference in the predicted mean strength for the two methods at 1% alpha level. b) Solve part (a) by Minitab and report the results.

Steel type	Method 1	Method 2
1	1.186	1.061
2	1.151	0.992
3	1.322	1.063
4	1.339	1.062
5	1.2	1.065
6	1.402	1.178
7	1.365	1.037
8	1.537	1.086
9	1.559	1.052

a) We wish to determine whether there is any difference in the predicted mean strength for the two methods at 1% alpha level.

b) Solve part (a) by Minitab and report the results.

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

	Method 1	Method 2	d = B - A	(d - dbar)^2
	1.186	1.061	0.125	0.022167901
	1.151	0.992	0.159	0.013199457
	1.322	1.063	0.259	0.000221679
	1.339	1.062	0.277	9.67901E-06
	1.2	1.065	0.135	0.019290123
	1.402	1.178	0.224	0.002488901
	1.365	1.037	0.328	0.002928012
	1.537	1.086	0.451	0.031368346
	1.559	1.052	0.507	0.05434079
Sum	12.061	9.596	2.465	0.146014889
Mean	1.340111111	1.066222222	0.273888889	0.016223877
St.Dev	13.51	15.89	13.22

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 0.1351

SE = s / sqrt(n)

S.E = 0.04503

DF = n - 1 = 9 -1

D.F = 8

t = [ (x₁ - x₂) - D ] / SE

t = 6.08

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 8 degrees of freedom is more extreme than 6.08; that is, less than - 6.08 or greater than 6.08

Thus, the P-value = 0.00

Interpret results. Since the P-value (0.000) is less than the significance level (0.01), we have to reject the null hypothesis.

Reject H0. We have sufficient evidence in the favor of the claim that there is difference in the predicted mean strength for the two methods at 1% alpha level

b)

Minitab output

Paired T-Test and CI: Method 1, Method 2

Descriptive Statistics

Sample	N	Mean	StDev	SE Mean
Method 1	9	1.3401	0.1460	0.0487
Method 2	9	1.0662	0.0494	0.0165

Estimation for Paired Difference

Mean	StDev	SE Mean	95% CI for μ_difference
0.2739	0.1351	0.0450	(0.1700, 0.3777)

µ_difference: mean of (Method 1 - Method 2)

Test

Null hypothesis	H₀: μ_difference = 0
Alternative hypothesis	H₁: μ_difference ≠ 0

T-Value	P-Value
6.08	0.000