Question

19 Two Suppliers produce the exact component for a particular Customer. Random samples were drawn from the last lot sent in by each Supplier to see if they had any significant difference. Supplier "A" had a mean of 1.607" with a standard deviation of.0020 from a sample of 10 parts. Supplier "B" had a mean of 1.606" with a standard deviation equal to 0018 from a sample of 10 parts. The Customer target value is 1.606". Supplier "B" charges $3.00 per part more than Supplier "A". Which recommendation below would you make to your boss? Use a 95% confidence level. (HINT: Use t-test for between two independent populations. First do F-Test to see if you need to use 2 Sample T-Test Assuming Equal Variances or 2 Sample T-Test Assuming Unequal Variances). Show all work or use Minitab/Excel. No work no credit 6) A) Go with Supplier "B" only since B hit target on nose with smaller variation. B) Go with either Supplier "A" or "B" since we could not reject the null hypothesis and conclude no significant difference between the two Suppliers C) Go with Supplier "A" as the primary Supplier and use "B" as a backup since we could not reject the null hypothesis and conclude no significant difference between the two Suppliers. Since they are not significantly different, we can save money going with Supplier "A" Get out of both Suppliers since neither are meeting the Customer requirement. D)

Answer 1

Solution: Test for the equality of variance: MINITAB procedure: Step 1: Choose Stat > Basic Statistics> 2 Variance Step 2: Under Data, choose Sample standard deviation. Step 3: In First, enter 10 under Sample size. Step 4: In First, enter 0.0020 under Standard deviation Step 5: In Second, enter 10 under Sample size. Step 6: In Second, enter 0.0018 under Standard deviation Step 7: Check Options, enter Confidence level as 95.0. Step 8: In Hypothesized ratio Var 1 / Var 2 Step 9: Choose not equal in alternative. Step 10: Click OK in all dialog boxes. MINITAB output:

Test and Cl for Two Variances Method Null hypothesis Alternative hypothesis Variance (1) Variance (2) not1 significance level Variance(1) Variance (2)1 0 . 05 Alpha Statistics Sample StDev Variance 0.000 0.000 10 0.002 10 0.002 Ratio of standard deviations 1.111 Ratio of variances 1.235 95% Confidence Intervals cI for Variance Ratio Distribution CI for StDev of Data Normal Ratio (0.554, 2.229) (0.307, 4.970) Tests Test DE1 DE2 Statistic P-Value 1.23 0.759 Method F Test (normal) From MINITAB output, the p-value is, p-value 0.759

Since p-value 0.05, thus, it can be concluded that the equality of variances can be assumed Test for two sample t test. Null hypothesis: Alternative hypothesis: MINITAB Procedure: Step 1: Choose Stat> Basic Statistics > 2-Sampleit Step 2: Choose Summarized data Step 3: In first, enter Sample size as 10, Mean as 1.607, Standard deviation as 0.002. Step 4: In second, enter Sample size as 10, Mean as 1.606, Standard deviation as 0.0018. Step 5: Select Assume equal variances. Step 6: Choose Options. Step 7: In Confidence level, enter 95 Step 8: In Alternative, select not equal. Step 9: Click OK in all the dialogue boxes. MINITAB output:

Two-Sample T-Test and Cl Sampleen StDey SE Mean 10 1.60700 0.00200 0.00063 10 1.60600 0.00180 0.00057 Difference#mu (1) -mu (2) Estimate for difference: 0.001000 95t CI for difference: -0.000788, 0.002788) T-Test of difference-0 (vs not =): T-value 1.18 p-value = o.255 DF = 18 Both use Pooled StDev = 0.0019 From the MINITAB output, the 95% confidence interval is, CI -0.00079,0.0028) Since, the 95% confidence interval contain the value 0 in it, therefore, it can be concluded that there is no difference in the means Therefore, the correct option is option (C).