CSUF ISDS 361A Practice Exam 2 Version B Skordi Questions and Answers Verified 100% Correct

CSUF ISDS 361A Practice Exam 2 Version B Skordi Questions and Answers Verified 100% Correct The following 5 questions are based on this information. Historically, the average number of boats owned by the people living in the coasts in a lifetime has been 12. An economist believes that the number is now lower because of recent economic downturns. A recent survey of 30 senior citizens indicates that the average number of boats owned over their lifetime is 10. Assume that the random variable, number of boats owned in a lifetime (denoted by X), is normally distributed with a standard deviation (σ) is 4.5. - ANSWER -You need to know these for the five questions Specify the null and alternative hypotheses. Select one: a. H(0): μ≥12 versus H(a): μ12 b. H(0): μ≤12 versus H(a): μ12 - ANSWER -an economist believes that the number is now lower. Hence, H(0): μ≥12 versus H(a): μ12 a. H(0): μ≥12 versus H(a): μ12 The standard error (SE) of X¯ is Select one: a. 2.5 b. 4.5 c. 10.5 d. 0.82 - ANSWER -SE = sigma/sqrt(n) n 30 x bar 10 sigma 4.5 d. 0.82 The p-value is Select one: a. 0.00 b. 0.03 c. 0.99 d. 0.05 - ANSWER -p value = NORM.S.DIST(Z obs,TRUE) a. 0.00 The test statistics value is Select one: a. -2.5 b. 2.5 c. 0.9 d. -2.43 - ANSWER -n 30 x bar 10 sigma 4.5 mean 0 12 SE 0.9 z obs -2.43 Z obs = x bar - mean 0/SE d. -2.43 At α=0.10 and using the p-value Select one: a. We do not reject H(0) b. We reject H(0) in favor of H(a) - ANSWER -As P value alpha. We reject H0. b. We reject H(0) in favor of H(a) The following 5 questions are based on this information. In a poll of 500 Graduat students, .75% (p¯=0.75) said that they used only Internet for the project and assignment purposes.. The goal is to construct a 99% confidence interval for the percentage (p) of Graduatel students who use the Internet for project and assignment purposes. - ANSWER -You need to know this for the next five questions The standard error (SE) of p¯ is Select one: a. 0.0004 b. 0.016 c. 0.0002 d. 0.019 - ANSWER -p bar 0.75 n 500 SE 0.019365 alpha 0.01 SE= =SQRT((p bar*(1- pbar))/n d. 0.019 The critical value (CV) needed for 99% confidence interval estimation is Select one: a. 2.58 b. 1.28 c. 1.64 d. 1.96 - ANSWER -P value =-NORM.S.INV(0.01/2) a. 2.58 The 99% confidence interval estimate of p is Select one: a. 0.44 ± 0.002 b. 0.75 ± 0.05 c. 0.75 ± 0.15 d. 0.44 ± 0.03 - ANSWER -p bar 0.75 n 500 alpha 0.01 SE 0.019365 CV 2.575829 ME 0.049881 ME = CV*SE IE= pbar +/- ME = 0.75 +/- 0.05 b. 0.75 ± 0.05 Suppose around the period the above poll was conducted,The DEan of a university made a personal statement saying that .85% of Graduate students used only the Internet for assignment purposes In light of the sample evidence and at the 1% level of significance, Select one: a. We cannot reject the Dean's claim b. We can reject the Dean's claim - ANSWER -Because 0.85 is not in the range 0.75 +/- 0.05. We can reject the claim. b. We can reject the Dean's claim A Dean of the universityl wishes to collect new random sample with the aim of building a new confidence interval at the 99% confidence level for p. Using the current sample proportion (from the 500 graduate students poll ) as a basis, what sample size (n) would the journalist require to achieve a 10% margin of error? Select one: a. 73 b. 500 c. 125 d. 250 - ANSWER -n= z_(α/2) ^2 * p * (1-p)/E^2 c. 125 The following 5 questions are based on this information. A survey indicates that the proportion of girls in the total youth in the United States is 51% (p=0.51). We will take a random sample of 400 U.S. youths. - ANSWER -You need to know this for the next five questions The sampling distribution of p¯, the sample proportion of U.S.youths who are girls, is: Select one: a. is normal because np≥5 and n(1−p)≥5