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(Solved): Question 13 9 pts Two different analytical tests can be used to determine the impurity level in ...



Question 13
9 pts
Two different analytical tests can be used to determine the impurity level in steel alloys. Eight specimens

Output 5.1:
Two-Sample T-Test and Cl: Test 1, Test 2
Two-sample \( T \) for Test 1 vs Test 2
Difference \( = \) mu (Test 1) \

Question 13 9 pts Two different analytical tests can be used to determine the impurity level in steel alloys. Eight specimens are tested using both procedures, and the results are shown below. Use the Minitab output provided to answer the following questions. Part ONE What is the test statistic and associated \( p \)-value for the test to determine whether the data is normally distributed (give values to three decimal places-no room for error)? Test Statistics: , \( \mathrm{p} \)-value: Are the data normally distributed (YES) or not normally distributed (NO)? Part TWO Test whether the mean level of impurity for both of the types of alloy are the same or not. State your hypotheses, a p-value, the decision made and your conclusion. Also indicate which output you used to make your decision. Select your hypothesis test pair (A,B or C): A \( H_{0}: \mu_{1}-\mu_{2}=0, H_{a}=\mu_{1}-\mu_{2} \neq 0 \) B \( H_{0}: \mu_{1}-\mu_{2}=0, H_{a}=\mu_{1}-\mu_{2}>0 \) ? \( H_{0}: \mu_{1}-\mu_{2}=0, H_{a}=\mu_{1}-\mu_{2}<0 \) From the output given, what is the correct test statistic? What is associated p-value? Which of the following is the appropriate conclusion (i or ii): i There is evidence to suggest that the the mean level of impurity for both of the types of alloy are the same. ii There is evidence to suggest that the the mean level of impurity for both of the types of alloy are not the same. Output 5.1: Two-Sample T-Test and Cl: Test 1, Test 2 Two-sample \( T \) for Test 1 vs Test 2 Difference \( = \) mu (Test 1) \( - \) mu \( ( \) Test 2\( ) \) Estimate for difference: \( =0.212 \) \( 95 \% \mathrm{Cl} \) for difference: \( (-0.483,0.058) \) T-Test of difference \( =0 \) (v5 not \( =): \mathrm{T} \)-Value \( =-1.70 \mathrm{P} \)-Value \( =0.114 \mathrm{DF}=13 \) Output 5.2: Output 5.3: Paired T-Test and Cl: Test 1, Test 2 Paired T for Test 1 - Test 2 \( 95 \% \mathrm{Cl} \) for mean difference: \( (-0.3636,-0.0614) \) T-Test of mean difference \( =0(v 5 \) not \( =0) \) : \( T \)-Value \( =-3.32 \) P-Value \( =0.01 \)


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Part (1) As per the MINITAB output be It is paired t-tes
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