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## Week 2 Electronic Text MicroSoft Word Solutions (9.53+9.57) - All work shown + megastat output

by: smartwriter Notetaker

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# Week 2 Electronic Text MicroSoft Word Solutions (9.53+9.57) - All work shown + megastat output

Marketplace > Week 2 Electronic Text MicroSoft Word Solutions 9 53 9 57 All work shown megastat output
smartwriter Notetaker
CSU - Dominguez hills
GPA 3.0

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Week 2 Electronic Text MicroSoft Word Solutions (9.53+9.57) - All work shown + megastat output
COURSE
PROF.
No professor available
TYPE
Study Guide
PAGES
5
WORDS
KARMA
50 ?

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Date Created: 11/15/15
9.53 On eight Friday quizzes, Bob received scores of 80, 85, 95, 92, 89, 84, 90, 92. He tells Prof. Hardtack that he is really a 90+ performer but this sample just happened to fall below his true performance level. (a) State an appropriate pair of hypotheses. (b) State the formula for the test statistic and show your decision rule using the 1 percent level of significance. (c) Carry out the test. Show your work. (d) What assumptions are required? (e) Use Excel to find the p­value and interpret it (a) H 0μ≥90 H 1μ<90 (b) x−−μ Test statistic(t)= ¯ s n √ Degreesof freedom=n−1=8−1=7 α=0.01 Criticalvalue=t0.01ith d . f .7=−2.998 Decisionrule:Reject thenull hypothesisif t<−2.998 Accept thenullhypothesis if t>−2.998 (c) x−¿¯ ¿ Samplemean¿ Standard deviatis)=4.984 μ=90 n=8 x−−μ¯ 88.375−90 −1.625 t= s = 4.984 = 1.762110099=−.9221898228≈−0.922 √n √8 We fail¿rejectthenullhypothesissinc e ­0.922 is not less than ­2.998. (d)  We must assume the quizzes are of the same difficulty, the same time limit is gave, the use of  calculator or no use of it is the same each time, and that the scores are normally distributed.  (e) P−value:P(t<−0.922)withdegreesof freedomequal¿7=0.1936 We fail to reject the null hypothesis since the p­value is not less than 0.01. We don't have  sufficient evidence to conclude that Bob is not a 90+ performer. 9.57 A sample of 100 one­dollar bills from the Subway cash register revealed that 16 had something written on them besides the normal printing (e.g., “Bob ♥ Mary”). (a) At α = .05, is this sample evidence consistent with the hypothesis that 10 percent or fewer of all dollar bills have anything written on them besides the normal printing? Include a sketch of your decision rule and show all calculations. (b) Is your decision sensitive to the choice of α? (c) Find the p­value. H0 is 10% or fewer have writing Ha is greater than 10% have writing H 0p≤0.1 H 1p>0.1 z0.1.282 Reject thenullhypothesis isz>1.282 Otherwiseacceptthealternatehypothesis Compute the test statistic: 16 ¿ 100=0.16 p=.1 n=100 z= 0.16−0.1) 0.11−0.1) [ 100 ] √ 0.06 z=√0.0009 z=2 Since 2>1.282, we reject the null hypothesis.  P(z>2=0.0228 The p value (from a table) is 0.0228. This is less than a = 0.05, so reject the null hypothesis.  There seems to be more than 10% of bills with writing. Yes, this is sensitive to alpha. If alpha was, for example, 0.01, we would not reject the null  hypothesis.

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