A Harris corporation/University of Florida study was undertaken to determine whe

Question

A Harris corporation/University of Florida study was undertaken to determine whether a manufacturing process performed at a remote location can be established locally. Test devices were set up at both the old and new locations and voltage readings on the process were obtained. The table below contains voltage readings from the two locations. Construct a 99% confidence interval for estimating the ratio between the two population variances. Is it reasonable to assume equal variances? Based on your conclusion above, is there a significant difference between the two population averages? Use = 0.01.

Explanation / Answer

Here we have to find 99% confidence interval for estimating the ratio between the two population variances.

Also we have to test the hypothesis that,

H0 : Variances are equal

H1 : Variances are not equal.

Assume alpha = level of significance = 5% = 0.05

Test statistic follows F-distribution.

We can do this test is MINITAB.

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 2 Variaces --> samples in different columns --> First : select old location --> Second : Select new location --> Options --> COnfidence level : 95.0 --> ok --> Storage : Click on variances --> ok --> ok

Test for Equal Variances

Level1 old location
Level2 new location
ConfLvl 95.0000

Bonferroni confidence intervals for standard deviations

Lower Sigma Upper N Factor Levels

0.164585 0.219193 0.323964 24 old location
0.371236 0.487098 0.700637 27 new location

F-Test (normal distribution)

Test Statistic: 0.202
P-Value : 0.000

Here P-value < alpha

Reject H0 at 5% level of significance.

Conclusion :  Variances are not equal.

We have given 95% confidence interval for ratio of standard deviations.

95% confidence interval for ratio of standard deviations is (0.164585, 0.323964)

95% confidence interval for ratio of variances is (0.027, 0.105)

Now we have to test the hypothesis that,

H0 : mu1 = mu2 Vs H1 : mu1 not= mu2

where mu1 and mu2 are two populations means.

Assume alpha = level of significance = 0.01

Here test statistic follows two sample t-distribution.

We can do two sample t-test in MINITAB.

steps :

ENTER data into MINITAB sheet --> STat --> Basic statistics --> 2-Sample t --> Samples in different columns --> First : old location --> Second :new location --> Do not click on assume equal variances --> Options -->Confidence level : 99.0 --> Test difference : 0.0 --> ALternative : not equal --> ok --> ok

Two-Sample T-Test and CI: old location, new location

Two-sample T for old location vs new location

N Mean StDev SE Mean
old loca 24 9.998 0.219 0.045
new loca 27 9.445 0.487 0.094

Difference = mu old location - mu new location
Estimate for difference: 0.552
99% CI for difference: (0.270, 0.834)
T-Test of difference = 0 (vs not =): T-Value = 5.32 P-Value = 0.000 DF = 37

Test statistic = 5.32

P-value = 0.000

P-value < alpha

Reject H0 at 6% level of significance.

Conlusion : There is sufficient evidence to say that the two population means are not equal.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.