Question

Textile manufacturer, Joywill International, is trying to decide which vendor they should source for a yarn needed in one of their products. Vendor A has a lower price than Vendor B but the determining factor must be the mean breaking strength of the yarn.

Random samples were drawn from the vendors’ supply, with results shown below:

	Vendor A	Vendor B
Sample size	10	12
Sample mean	105	98
Sample variance	8	9

Do these data provide sufficient evidence that the mean yarn strength for Vendor A is the more than the mean yarn strength for Vendor B.

Perform the appropriate hypothesis test using a significance level of 0.05.

Do your work in an Excel file. Make Excel do the test statistic calculation.!!!!!!

Answer 1

using excel

we have

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.05
Vendor A
Sample Size	10
Sample Mean	105
Sample Standard Deviation	2.828427
Vendor B
Sample Size	12
Sample Mean	98
Sample Standard Deviation	3		Formula used
Intermediate Calculations
Population 1 Sample Degrees of Freedom	9		B7-1
Population 2 Sample Degrees of Freedom	11		B11-1
Total Degrees of Freedom	20		B16+B17
Pooled Variance	8.5500		((B16*B9^2)+(B17*B13^2))/B18
Standard Error	1.2520		SQRT(B19*(1/B7+1/B11))
Difference in Sample Means	7.0000		B8-B12
t Test Statistic	5.5911		(B21-B4)/B20
Upper-Tail Test
Upper Critical Value	1.7247		(TINV(2*B5,B18))
p-Value	0.0000		TDIST(ABS(B22),B18,1)
Reject the null hypothesis

since p-value of test statistic t is 0.000 which is less than 0.05 so we will not reject Ho and conclude that we have sufficient evidence that the mean yarn strength for Vendor A is the more than the mean yarn strength for Vendor B.