All answers must include: 1. the rationale of using a test, 5. the p-value 6. th
ID: 3316334 • Letter: A
Question
All answers must include: 1. the rationale of using a test, 5. the p-value 6. the test result, 7. the distribution chart, and 8. the conclusion. 2. the hy potheses, 3. the critical-value 4. the test-statisties The manufacturer of hardness testing equipment uses steel-ball indenters to penetrate metal that is being tested. However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made. Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a 95% confidence interval and test whether the two indenters result in different measurements. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. 4 Specimen Steel ball Diamond 53 63 74 69 56 68 51 56Explanation / Answer
The rationale of this test is to see if there is a difference between the means of the two samples drawn.
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (9-1)*7.57^2+(9-1)*8.17^2/8+8
= 400.75+467.44/16
= 54.26
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(59-60.56)-0/54.26(1/9+1/9)
=-1.56/3.47
=-0.448
tCRIT is +/-2.12 and hence cannot reject the null hypothesis
Confidence interval:-
(X1-X2)+-ta/2*Sp^2(1/n1+1/n2)
-1.56+/-2.12*3.47
LCL=-1.56-7.3564=-8.9164
UCL=-1.56+7.3564=5.7964
(-8.9164,5.7964)
Steel ball Diamond 51 53 57 55 61 63 71 74 68 69 54 56 65 68 51 51 53 56 59 60.55555556 Average 7.566372975 8.171767115 Std DevRelated Questions
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