Question

Analysis of a random sample of 15 specimens of cold-rolled steel to determine yield strengths resulted in a sample mean strength of 29. 8 and a standard deviation of 4. 0. A second random sample of 14 specimens of galvanized steel resulted in a sample mean of 32. 7 and a standard deviation of 5. 0. Does the data indicate that the true mean yield strengths for the two given populations (cold-rolled or galvanized) are different? Test at a= 0. 01. Make sure to find/estimate the p-value.

Answer 1

Given:

Cold Rolled Steel: = 29.8, s1 = 4, n1 = 15

Galvanized Steel: = 32.7, s2 = 5, n2 = 14

Since s1 / s2 = 4 / 5 = 0.8 (between 0.5 and 2) we use the pooled variance

The degrees of freedom used = n1 + n2 - 2 = 15 + 14 - 2 = 27

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The Hypothesis:

H0:

Ha:

This is a Two tailed test.

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

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The p Value: The p value (2 Tail) for t = -1.72, df = 27, is; p value = 0.0969

The Critical Value:   The critical value (2 tail) at = 0.01 ,df = 27, tcritical = +2.77 and -2.77

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The Decision Rule: If tobservedis >tcritical or If tobserved is < -tcritical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision:    Since t lies in between +2.77 and -2.77, We Fail To Reject H0

Also since P value (0.0969) is > (0.01), We Fail to Reject H0.

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The Conclusion: There isn't sufficient evidence at the 99% significance level to support the claim that the true mean yield strengths for the 2 given populations (cold rolled and galvanized) are different.

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