Ten observations on etch uniformity on silicon wafers are taken during a qualifi
ID: 1718987 • Letter: T
Question
Ten observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: Construct a 95 percent confidence interval estimate of sigma^2. A historic distribution information of people visit on a Movie Theater is given as: Monday: 5%; Tuesday 5%; Wednesday: 10%; Thursday: 10%; Friday: 25%; Saturday: 25%; Sunday: 20%, and the total is 100%. A test is conducted by counting the number of people visits each day, and obtained as: Monday: 18; Tuesday 30; Wednesday: 34; Thursday: 24; Friday: 45; Saturday: 69; Sunday: 30. Based on the information, can you conclude with 95% confidence that the given distribution information is correct?Explanation / Answer
5) Mean = (5.34 + 6.65 +4.76 + 5.98 + 7.25 +6.00 +7.55 + 5.54 + 5.62 + 6.21) / 10 = 6.09
Std Deviation = sqrt( sum(Xi - Xavg)^2/10) = 0.817325
95 % confidence interval => 6.09 + 1.96 * 0.817325 = 7.691957
6.09 - 1.96 * 0.817325 = 4.488
interval is (4.488 , 7.691957)
6) Mean = (18 + 30 + 34 + 24 + 45 + 69 + 30) / 7 = 35.71429
Std deviation = 15.6453
Confidence interval (35.71429 + 1.96 * 15.6453 , 35.71429 - 1.96 * 15.6453)
( 66.379 , 5.0495)
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