Mechanical engineers have developed a new high-strength aluminum alloy for use i
ID: 3153349 • Letter: M
Question
Mechanical engineers have developed a new high-strength aluminum alloy for use in aircrafts, tankers, and long-range bombers. A series of strength tests were conducted to compare the new alloy to the current strongest alloy. In particular, three specimens of each type of aluminum alloy were produced and the yield strength (measured m MPa) of each specimen was determined. The results are summarized in the table below. Assuming identical population variances, construct a 99% confidence interval for the difference between the mean yield strengths of the two aloys. Does this confidence interval support the daim that the new aloy is superior to the current strongest alloy at 99% confidence level?Explanation / Answer
a)
Calculating the means of each group,
X1 = 631
X2 = 592
Calculating the standard deviations of each group,
s1 = 13.3
s2 = 12.4
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 3
n2 = sample size of group 2 = 3
Thus, df = n1 + n2 - 2 = 4
Also, sD = 10.49841258
For the 0.99 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.005
t(alpha/2) = 4.604094871
lower bound = [X1 - X2] - t(alpha/2) * sD = -9.33568751
upper bound = [X1 - X2] + t(alpha/2) * sD = 87.33568751
Thus, the confidence interval is
( -9.33568751 , 87.33568751 ) [ANSWER]
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b)
As 0 is inside this interval, then NO, THIS DOS NOT SUPPORT THE CLAIM. [ANSWER]
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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!