Confidence Interval and Hypothesis Test for Two Population Means Two suppliers m

Question

Confidence Interval and Hypothesis Test for Two Population Means Two suppliers manufacture a plastic gear used in a laser printer. The impact of these gears measured in foot-pounds is an important characteristic of interest. A random sample of 10 gears from Supplier 1 results in_1 = 296 and s_1 = 12. Another random sample of 25 gears from Supplier 2 results in_2 = 315 and s_2 = 24. Assume the populations are independent and normally distributed. Is there sufficient evidence to conclude that the variance of impact strength is different for the two suppliers at the = 0.05 level of significance? a) What is the parameter of interest? What assumptions are made? b) Write the null and alternative hypotheses. c) Calculate the test statistic. d) Determine the reject region. e) Make a decision and write a thorough interpretation in context of the problem. Is there evidence to support the claim that Supplier 2 provides gears with higher mean impact strength? Use = 0.05. a) Wha

Explanation / Answer

Solution:

We are given

X1bar = 296

X2bar = 315

S1 = 12

S2 = 24

N1 = 10

N2 = 25

Here, we have to use the F test for two population variances.

Parameters of interest are population variances of two populations. The assumption made that the given two populations are independent and normally distributed.

The null and alternative hypotheses are given as below:

H0: 1^2 = 2^2 versus Ha: 1^2 2^2

Test statistic formula is given as below:

F = S1^2 / S2^2 = 12^2/24^2 = 2304

First df = 24

Second df = 9

Upper critical value = 3.6142

P-value = 0.025

Alpha value = 0.05

P-value < Alpha value

Reject the null hypothesis that two population variances are same.

This means we conclude there is sufficient evidence that two population variances are different.

Now, we have to test

H0: µ1 = µ2 versus Ha: µ1 µ2

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

t = (296 – 315)/sqrt[(12^2/10)+(24^2/25)]

t = -3.1052

Lower critical value = -2.0935

Upper critical value = 2.0395

(By using t table or excel)

P-value = 0.0040

Alpha value = 0.05

P-value < Alpha value

So, we reject the null hypothesis that two population means are same.

We conclude that there is sufficient evidence that the two population means are different.

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