Question

need help with 20-23

option Semester 1-2016 19. An article poblished in the Canadian Journal of Zoology preseneed a method for estimacing the 25 fat percentage of North Amerian poroupines. The method was illustrated with a sample of n body fa pompees Based on this sample, a 95% bootstrap codence interval for the nerage be ntage of poecuptes is17 4% to 2S8% which ofthe following nul hypotheses would he rejected based on this confidence interval? 18.6%. A) H0 : tonua C) Ho : μ = 20.0% D) He : 22.9% E) Ho μ-24.6%, Scenario 4 Admi ions records at MIT indicates that 6,7% of the graduate students enrolled are from 20. Refer to Scenario 4. What is the minimum sample size for which the Central Limit Theorem applies in this case? 30, B) n=40, C) n=so. D) n 100 E) n 200 to Scenario 4. Find the mean and standard error of the sample proportion of Canadian students in random samples of 100 graduate students at MITT p-0.067, SE-00625. A) n冫100 B) p C 0.067, SE-0.025 0670, SE =0.250. 0067, s E-0.006. C.? D) (E) β-0.067, SE-0.0057 lo 22. Refer to Scenario 4. Roughly what percentage of samples of 100 randomly selected graduate at MIT will have at least 10% of students from Canada? A)5% 6.7% C) 10%. D) 18%. E) 25%, 23. For a N(o, 1) density what is the area to the left of s -1.645. A)2.5%. B)3.5%. 05%. D)1096. E) 11%. Downloaded by Thanh Tran ([email protected]

Answer 1

19. B) is correct as Ho: u = 26.6% falls outside the confidence interval. Hence, it would be rejected.

20. For central limit theorem, sample size should be greater than 30.
A) is correct.

21. p = 0.067
SE = sqrt(p(1-p)/n) = 0.025
Hence, C) is the correct answer.

22. Mean = u = 0.067
Standard deviation = sqrt(npq) = 0.025

P(X>0.1) = P(z>(0.1-0.067)/0.025) = P(z>1.32) = 1 - 0.9066 = 0.0934
Hence, C) 10% is correct

23. P(z<-1.645) = 0.05
Hence, C) is correct