Homework%208.pdf (1)
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School
University of Minnesota-Twin Cities**We aren't endorsed by this school
Course
STAT 3011
Subject
Statistics
Date
Dec 17, 2024
Pages
4
Uploaded by UltraApeMaster1061
1- D2- C3-B4- C5-CPart II- 1 - B-for 99% ∝=0.01-P value > 0.01 So we fail to reject Ho . Therefore 99 % confidence interval will always contain 0 2- -A-H0: P=0.79-Ha:=P<0.79-Left tailed test -B-Type 1 error: concluding P.0.70 when ≤0.79 -Means the Researcher Concludes that the Proportion of Accounting Firms Offering Flexible Scheduling is Greater than 79%, but in reality,It is Not-Type 2 error: failing to conclude p>0.70 when p>0.79-It Means the Researcher Fails to Conclude that the Proportion of Accounting Firms Offering Flexible Scheduling is Greater than 79% Even though It is Actually Large-C-Test statics formula -z= (p̂-po)/ (√ po(1-po)/n)-Z=(1-0.79)/(√ 0.79(1-0.79)/-P-value -P-value= Pnorm(x,Mean,Sd)-Za=1.645-Decision -If Z>Za= reject H0 otherwise , fail to reject H0-Siince Z< 1.645, we fail to reject H0-Conclusion -Insufficient Evidence to Prove p>0.79 For Testing If p<0.79 Hypotheses and Rejection Region would Adjust Accordingly3-A-Assumptions -Random samples: The samples from machines A and B are assumed to be random.
-Independence: Each sampled part's quality is independent of the others.-Large enough sample size for normal approximation: Since both sample sizes are greater than 30, we can use a normal approximation.-Null and alternative hypotheses -H0; PB=PA-The proportion of defective parts for machine B (PB) is not greater than the proportion for machine A (PA).-HA: PB≥PA-The proportion of defective parts for machine B is greater than the proportion for machine A.-Test statistic -Z= (p1-p2)-0/√ p(1-p)(1/n1=1/n2)-Z= ((22/70)-(5/35))-0/√ (27/105)(1-(27/105))(1/(1/70)=1/(1/35))-z= 0.3142-0.1429/√0.0028-Z=1.894662-P-value -ince the alternative hypothesis is HA: PB > PA , we are looking for the upper tail probability.-p-value = pnorm(z)-0.0290686-Conclusion-With a p-value of approximately 0.0290686 is grather than Alpha, we Fail to reject the null hypothesis. There is enough evidence to conclude that the proportion of defective parts for machine B is greater than the proportion for machine A at a 0.02 level of significance-Error type -Since we reject the null hypothesis, we are making a Type I error if the null hypothesis is actually true -B
--prop.test(c(5,22),c(35,70), alternative ="greater")-C-The conclusion will remain the same (reject the null hypothesis)-The p-value will still be much smaller than 0.05. Therefore, we would still conclude that the proportion of defective parts for machine B is greater than the proportion for machine A.-Rejecting a true null hypothesis is a Type I error. In this case, if the null hypothesis is true (machine B is not actually more defective), but we conclude that it is, we've made a Type I error.-D -(Pb-Pa)土Z √Pb(1-pb)/nb+Pa(1-Pa)/Na-We would use P hat for this -Cl=(46/100-11/35)土1.645√(47/100-(1-47/100)/100)+ (11/35(1-1135/35) -Cl=0.3566土0.1791-0.1773 to 0.5355-E
--prop.test(c(47,11),c(100,35), alternative ="two.sided", conf.level =0.90)-Result provides CI: (0.1773,0.5355)-F-Since we already have the CI for Pb-Pa, we can simply reverse the sign to get the CI for Pa-Pb-Cl(Pa-Pb)= (-0.5355,-.01773)-This covers all the parts of your question regarding hypothesis testing, confidence interval, and performing the analysis in R