ISDS Mock Exam 2 notes

.docx

School

California State University, Fullerton *

*We aren’t endorsed by this school

Course

361A

Subject

Statistics

Date

May 3, 2024

Type

docx

Pages

Uploaded by ChancellorBisonMaster1055 on coursehero.com

ISDS Mock Exam 2 notes A. The following 6 questions are based on this information. According to a report in U.S. News from April 2013, the average price of gasoline at all U.S. gas stations was $3.00 (µ). The population standard deviation (σ) of gas prices is $1.75. Let X be a random variable denoting gas price at gas stations. We plan to take a random sample of 36 gas stations. 1 . CLT n > 30 2 . If X is normal, a sample of Xbar = 16 is normal. 3 . What is the probability that a random sample of 36 gas stations will provide an average gas price ( 𝑋 bar) that is more than $3.50? P (Xbar>3.50) = P(Z>1.7143) = 3.5-3/(1.75/SQRT(36)) = 1.7143 Find the probability of Z. P(Z>1.7143) = 1-P(Z<1.7143) = 1 – Norm.S.Dist (1.7143,true) = 4.3% 4 . What is the probability that a random sample of 36 gas stations will provide an average gas price ( 𝑋 bar) that is within $0.50 of the population mean ( 𝜇 )? Find P(Xbar-Mu < 0.5) = P (-0.5 < Xbar-Mu < 0.5) = P (-0.5+Mu < Xbar < 0.5 + Mu) = P(-0.5+3 < Xbar < 0.5+3) = P(2.5 < Xbar < 3.5) Then transform to standard normal P [ (2.5-3)/1.75/SRQT(36) < Z < (3.5-3)/1.75/SQRT(36)) ] = P(-1.7<Z<1.7) = P(Z<1.7) – P(Z<-1.7) = Norm.S.Dist(1.7,true) – Norm.S.Dist(-1.7,true) = 91% 5 . The probability in the PREVIOUS question would INCREASE if we were to increase the sample size to 72. B . The following 5 questions are based on this information. Census indicates that the proportion of adult women in the United States is 51% ( 𝑝 =0.51). We will take a random sample of 400 U.S. adults. 6 . The sampling distribution of 𝑝 bar, the sample proportion of U.S. adult who are women, is: is normal because 𝑛𝑝 ≥5 and 𝑛 (1− 𝑝 )≥5. 7 . Find Standard error. P=0.51, n = 400. SE = SQRT ((P*(1-p))/n) = 0.025 8 . What is the probability (rounded) that a random sample of 400 U.S. adults will provide a sample proportion ( 𝑝 bar) that is within 0.03 of the population proportion ( 𝑝 )? SE = SQRT ((P*(1-p))/n) = 0.025. Find Z -0.03/0.025 = P(-1.2 < Z < 1.2) =NORM.S.DIST(1.2,TRUE)- NORM.S.DIST(-1.2,TRUE) = 0.77 9 . What is the probability (rounded) that a random sample of 400 U.S. adults will provide a sample proportion ( 𝑝 bar) that is within 0.07 of the population proportion ( 𝑝 )?

SE = 0.025. Find Z -.07/0.025 = -2.8 = P(-2.8<Z<2.8) =NORM.S.DIST(2.8,TRUE)-NORM.S.DIST(-2.8,1) = 0.99. 10 . Say you took a random sample of 400 U.S. adults and found out that the sample proportion ( 𝑝 bar) for this sample is 0.42. This is a rare finding because the likelihood of 𝑝 bar =0.42 is quite large as we saw in the previous question. Sample is too large to be a rare finding. C . A sample of n=25 cereal boxes of Granola Crunch, a generic brand of cereal, yields a mean ( 𝑋 bar) weight of 1.02 pounds. The goal is to construct a 95% confidence interval for the mean weight (µ) of all cereal boxes of Granola Crunch. Assume that the weight of cereal boxes is normally distributed with a population standard deviation (σ) of 0.03 pounds. 11 . The Standard error (SE) of Xbar is. SE= SDx = 0.03/SQRT(25) = 0.06 12 . The critical value (CV) used for a 95% interval estimate is Z(1-(.05/2)) = .975 =NORM.S.INV(0.975) = 1.96 13 . The 95% confidence interval estimate of µ is. Mu = Xbar +- Zcv X SDxbar = 1.02 +- 1.96 x .06 = 1.02 +- 0.012 14 . Suppose you think that average weight of a cereal box of Granola Crunch is 0.9 pounds. In light of the sample evidence and at the 5% level of significance. Your claim is not statistically justified as the range is 1.02 +- .012 15 . If we increase the confidence level (1-α) from 0.95 to 0.99, the margin of error (ME) of the confidence interval estimate will. Increase D . Based on a random sample of 30 U.S. workers, a researcher found that the average ( 𝑋 bar) number of vacation days a U.S. worker takes per year is 16. The sample standard deviation ( s ) is 12 days. The goal is to construct a 99% confidence interval for the average ( 𝜇 ) number of vacation days taken by U.S. workers per year. Assume that the number vacation days taken by U.S. workers is normally distributed. 16 . Standard error (SE) of Xbar = s/SQRT(n) = 12/(30) = 2.19 17 . The critical value (CV) needed for 99% confidence interval estimation is = alpha = .01, Alpa/2 = .005, T.INV(0.005,29) = 2.76 18 . The 99% confidence interval estimate of 𝜇 is? Confidence interval (CI) = mean +/- t*SE = 16 +/- 2.76*2.19 = 16+/- 6.04

19 . Suppose CEO of a company claims that the yearly average vacation days of all U.S. workers is 18 days. In light of the sample evidence and at the 1% level of significance. We cannot reject the CEO's claim. 20 . If we decrease the confidence level (1-α) from 0.99 to 0.95, the margin of error (ME) of the confidence interval estimate will decrease . F . Last year, a study showed that the average ATM cash withdrawal took 65 seconds with a standard deviation of 10 seconds. The study is to be repeated this year, but we need to determine the required sample size (n). Then, you will use a random sample from this year to construct a confidence interval for the average (mean) time it takes to withdraw cash from an ATM. 21 . For the confidence level (1 - α) = 0.95, the critical value (CV) is alpha = .05/2 = .025 Z test = NORM.S.INV(0.025) = 1.96. CV=1.96 22 . How large a sample (rounded up) would be needed to estimate this year's mean with 95% confidence and a margin of error of 4 seconds? Use the standard deviation from last year study as the population standard deviation (σ). round up 23 . How large a sample (rounded up) would be needed to estimate this year's mean with a 95% confidence and a margin of error of 2 seconds? Use last year's standard deviation as the population standard deviation (σ). 24 . For the confidence level (1-α) = 0.99, the critical value (CV) is (1 - α) = 0.99; α = 1-0.99 = 0.01 ; α/2 =0.01/2 =0.005, NORM.S.INV(0.005) = 2.58 the CV=2.58 25 . How large a sample (rounded up) would be needed to estimate this year's mean with a 99% confidence and a margin of error (ME) of 4 seconds? Use the standard deviation from last year as the population standard deviation (σ). In a poll of 1000 American adults conducted on August 5-9, 2010, 44% ( 𝑝 bar=0.44) of respondents approve the job that President Barack Obama was doing handling the economy. The goal is to construct a 90% confidence interval for the percentage ( 𝑝 ) of American adults who approved of Barack Obama's handling of the economy around the period the poll was conducted. 26. The standard error (SE) of 𝑝 ¯ is

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version