Question

Find and Insert e object th Replace Select all Editing lisplayed improperly. 23. 4. 57 • A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It is important that the diameters be as close as possible to an industry standard. The output from each process is sampled and the average error from the industry standard is calculated. The results are presented below. (20 marks) Process Process D Mean 0.005mm 0.0066mm Standard Deviation 0.0001mm 0.00012mm Sample Size 15 16 The researcher is interested in determining whether there is evidence that the two processes yield different average errors. Given the following excel printout, what analysis and decision can be made? Process 0.005mm 0.0001 15 Hypothesis Test: Independent Groups Process D 0.0066mm mean 0.00012 std dev. 16 desen Use these results to complete the following hypothesis test to answer the question of interest. (8 marks) We are interested in the hypotheses: Ho: vs HA where (define any parameters used above) So, at a significance level of a = 0.04, we would (circle one): (4 marks) accept the null hypothesis / reject the null hypothesis / fail to reject the null hypothesis and conclude that give your conclusion in the context of the problem of interest t)

Answer 1

Solution :

The null and alternative hypotheses are as follows :

Ho: Mc = HD

$\large H_{a}: \mu_{C}\neq\mu_{D}$

Where, are population mean errors for process C and process D respectively.

Вс, Др

To test the hypothesis we shall use the two-independent samples t-test. The test statistic is given as follows :

$\large t = \frac{(\bar{x}_{c}-\bar{x}_{D})}{\sqrt{S^{2}_{pooled}\left (\frac{1}{n_{1}}+\frac{1}{n_{2}}\right )}}$

Where, $\large \bar{x}_{C}, \bar{x}_{D}$ are sample means, $\large n_{1},n_{2}$ are sample sizes.

$\large S^{2}_{pooled}=\frac{(n_{1}-1)(s_{C})^{2}+(n_{2}-1)(s_{D})^{2} }{n_{1}+n_{2}-2}$

$\large s_{C}, s_{D}$ are sample standard deviations.

We have, $\large s_{C} = 0.0001,s_{D}=0.00012,n_{1}=15,n_{2}=16$

$\large S^{2}_{pooled}=\frac{(15-1)(0.0001)^{2}+(16-1)(0.00012)^{2}}{15+16-2}$

$\large S^{2}_{pooled}=0.00000001228$

$\large \therefore t = \frac{(0.005-0.0066)}{\sqrt{0.00000001228\left ( \frac{1}{15}+\frac{1}{16} \right )}}=-40.1808$

The value of the test statistic is -40.1808.

Degrees of freedom = (n1 + n2 - 2) = (15 + 16 - 2) = 29

Since, our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic. The two-tailed p-value is given as follows :

p-value = 2.P(T > |t|)

We have, |t| = 40.1808

p-value = 2.P(T > 40.1808)

p-value = 0.0000

The p-value is 0.0000.

Significance level = 0.04

(0.0000 < 0.04)

Since, p-value is less than the significance level of 0.04, therefore we shall reject the null hypothesis at 0.04 significance level.

Conclusion : At significance level of 0.04, there is sufficient evidence to conclude that the two process yield different average errors.

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