Question

Testing the Difference Between Two Means In Exercises 13–22, (a) identify the claim and state H0 and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic t, (d) decide whether to reject or fail to reject the null hypothesis and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.

19. Tensile Strength The tensile strength of a metal is a measure of its ability to resist tearing when it is pulled lengthwise. An experimental method of treatment produced steel bars with the tensile strengths (in newtons per square millimeter) listed below.

Experimental Method:
391 383 333 378 368 401 339 376 366 348
The conventional method produced steel bars with the tensile strengths
(in newtons per square millimeter) listed below.

Conventional Method:
362 382 368 398 381 391 400
410 396 411 385 385 395 371

At a = 0.01, can you support the claim that the experimental method of treatment makes a difference in the tensile strength of steel bars? Assume the population variances are equal.

Answer 1

The following is obtained from the given data

	Experimental	Conventional
Total	3683	5435
n	10	14
Mean	368.30	388.21
SS	4476.1	2846.3574
Var	497.3444	218.9506
SD	22.301	14.797

Since we assume population variance to be equal, we use the pooled variance

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(a) The Hypothesis:

H0:

Ha: : Claim

This is a Two tailed test

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(b) The Critical Value:   The critical value (2 tail) at = 0.01 ,df = 22, tcritical = +2.819 and -2.819

The Decision Rule: If t observed is > 2.819 or If t observed is < -2.819, Then Reject H0.

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(c) The Test Statistic: We use the students t test as population standard deviations are unknown.

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(d) The Decision: Since t lies in between +2.819 and -2.819, We Fail To Reject H0

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(e) The Conclusion: There isn't sufficient evidence at the 99% significance level to support the claim that the experimental method of treatment makes a difference in the tensile strength of steel bars.

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

Experimental					Conventional
#	X	X - Mean	(X - Mean)2		#	X	X - Mean	(X - Mean)2
1	391	368.3	515.29		1	362	388.21	686.9641
2	383	368.3	216.09		2	382	388.21	38.5641
3	333	368.3	1246.09		3	368	388.21	408.4441
4	378	368.3	94.09		4	398	388.21	95.8441
5	368	368.3	0.09		5	381	388.21	51.9841
6	401	368.3	1069.29		6	391	388.21	7.7841
7	339	368.3	858.49		7	400	388.21	139.0041
8	376	368.3	59.29		8	410	388.21	474.8041
9	366	368.3	5.29		9	396	388.21	60.6841
10	348	368.3	412.09		10	411	388.21	519.3841
					11	385	388.21	10.3041
					12	385	388.21	10.3041
					13	395	388.21	46.1041
					14	371	388.21	296.1841
Total	3683		4476.1000		Total	5435		2846.3574