Due in 4 hours. check the question before text me. make sure you really good at this.
\n\u00a0\u00a0
\nHypothesis Testing: Does Chance explain the Results?
\nHypothesis Tests
\nAn experimenter suspects that a certain die is \u201cloaded;\u201d that is, the chances that the die lands on different faces are not all equal. Recall that dice are made with the sum of the numbers of spots on opposite sides equal to 7: 1 and 6 are opposite each other, 2 and 5 are opposite each other, and 3 and 4 are opposite each other.
\nThe experimenter decides to test the null hypothesis that the die is fair against the alternative hypothesis that it is not fair, using the following test. The die will be rolled 50 times, independently. If the die lands with one spot showing 13 times or more, or 3 times or fewer, the null hypothesis will be rejected.
\nUnder the null hypothesis, the distribution of the number of times the die lands showing one spot is binomial with parameters n=50, p=1\/6\u00a0
\nUnder the null hypothesis, the expected number of times the die lands showing one spot is 8.3333 and the standard error of the number of times the die lands showing one spot is (Q6)\u00a0
\nProblem 4
\nThe significance level of this test is (Q7)\u00a0
\nProblem 5
\nThe power of this test against the alternative hypothesis that the chance the die lands with one spot showing is 29.28%, the chance the die lands with six spots showing is 4.05%, and the chances the die lands with two, three, four, or five spots showing each equal 1\/6, is (Q8)\u00a0
\nProblem 6
\nThe power of this test against the alternative hypothesis that the chance the die lands with two spots showing is 17.42%, the chance the die lands with five spots showing is 15.91%, and the chances the die lands with one, three, four, or six spots showing each equal 1\/6, is (Q9)\u00a0
\nSuppose that to have power against a wider variety of alternatives, the experimenter decides to base the test on the maximum number of times any face shows, instead of just the number of times one spot shows. That is, she will roll the die 50 times and calculate
\n(number of times die lands showing one spot) (number of times die lands showing two spots) (number of times die lands showing three spots) (number of times die lands showing four spots) (number of times die lands showing five spots) and (number of times die lands showing six spots).
\nShe will reject the null hypothesis if the largest (maximum) of those 6 random numbers is greater than 17.
\nProblem 7
\nUnder the null hypothesis, the distribution of the maximum number of times any face shows is (Q10) \u00a0\u00a0?\u00a0
\nA: geometric\u00a0
\nC: normal\u00a0
\nD: binomial\u00a0
\nE: negative binomial\u00a0
\nF: none of the above \u00a0
\nA manufacturer of computer memory chips produces chips in lots of 1000. If nothing has gone wrong in the manufacturing process, at most 7 chips each lot would be defective, but if something does go wrong, there could be far more defective chips. If something goes wrong with a given lot, they discard the entire lot. It would be prohibitively expensive to test every chip in every lot, so they want to make the decision of whether or not to discard a given lot on the basis of the number of defective chips in a simple random sample. They decide they can afford to test 100 chips from each lot. You are hired as their statistician.
\nThere is a tradeoff between the cost of eroneously discarding a good lot, and the cost of warranty claims if a bad lot is sold. The next few problems refer to this scenario.
\nProblem 8
\n(Continues previous problem.) A type I error occurs if a good lot is discarded\u00a0
\nProblem 9
\n(Continues previous problem.) A type II error occurs if a bad lot is not discarded\u00a0
\nProblem 10
\n(Continues previous problem.) Under the null hypothesis, the number of defective chips in a simple random sample of size 100 has a (Q14) \u00a0\u00a0?\u00a0
\nA: hypergeometric\u00a0
\nB: geometric\u00a0
\nC: binomial\u00a0
\nE: negative binomial\u00a0
\nF: none of the above \u00a0
\ndistribution, with parameters (Q15) \u00a0\u00a0?\u00a0
\nA: p=7\/1000\u00a0
\nB: mean=7\/1000 , SD=833.727\u00a0
\nC: N=1,000, G<\/p>\n","protected":false},"excerpt":{"rendered":"
Due in 4 hours. check the question before text me. make sure you really good at this. \u00a0\u00a0 Hypothesis Testing: Does Chance explain the Results?…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-102174","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/posts\/102174","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/comments?post=102174"}],"version-history":[{"count":0,"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/posts\/102174\/revisions"}],"wp:attachment":[{"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/media?parent=102174"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/categories?post=102174"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/academicwritersbay.com\/answers\/wp-json\/wp\/v2\/tags?post=102174"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}