Portland State University



Stat243BU Quiz #5 Version A Spring 2015 Students can use a page of note (front and back) prepared by themselves and a calculator. However, class notes and textbooks are not allowed. Each problem is of 5 points worth.Correct Answer in Red Explanations in Green1. A population has a standard deviation of 50. A random sample of 100 items from this population is selected. The sample mean is determined to be 600. At 95% confidence, the margin of error isa.5b.9.8c.650d.609.8ME=1.96*50/√(100)=9.82. The value added and subtracted from a point estimate in order to develop an interval estimate of the population parameter is known as thea.confidence levelb.margin of errorc.parameter estimated.interval estimate 3. Whenever the population standard deviation is unknown and the population has a normal or near-normal distribution, which distribution is used in developing an interval estimation?a.standard distributionb.z distributionc.alpha distributiond.t distribution 4. The t value for a 95% confidence interval estimation with 24 degrees of freedom isa.1.711b.2.064c.2.492d.2.069Row: 24 df, Column: .05 in two tails. OR invT(.975,24)=2.06395. In determining the sample size necessary to estimate a population proportion, which of the following information is not needed?a.the maximum margin of error that can be toleratedb.the confidence level requiredc.a preliminary estimate of the true population proportion pd.the mean of the populationn > pq(zα/2/E)26. A sample of 225 elements from a population with a population standard deviation of 75 is selected. The sample mean is 180. The 95% confidence interval for ? isa.105.0 to 225.0b.175.0 to 185.0c.100.0 to 200.0d.170.2 to 189.8180 ± 1.96*75/√(225)ME=9.87. Which of the following best describes the form of the sampling distribution of the sample proportion?a.When standardized, it is exactly the standard normal distribution.b.When standardized, it is the t distribution.c.It is approximately normal as long as n 30.d.It is approximately normal as long as np 5 and n(1-p) 5.8. In a random sample of 144 observations, p=0.6. The 95% confidence interval for p isa.0.52 to 0.68b.0.144 to 0.200c.0.60 to 0.70d.0.50 to 0.70.6 ± 1.96√(.6*.4/144)ME=.089 – 10. A new brand of chocolate bar is being market tested. Four hundred of the new chocolate bars were given to consumers to try. The consumers were asked whether they liked or disliked the chocolate bar. You are given their responses below.ResponseFrequencyLiked300Disliked1004009.Construct a 95% confidence interval for the true proportion of people who liked the chocolate bar.ANS: p=300400=0.75; α = 0.05; 0.71 to 0.79p±zα2p1-pn0.75 ±1.96(0.75)(1-0.75)400 0.75 ±1.96(0.022) (0.708, 0.792)10.With a .95 probability, how large of a sample needs to be taken to provide a margin of error of 3% or less?ANS:801n=(1.96)2(.75)(1-.75)(.03)2= .7203.0009=800.333 We round this up to 801.11. What is the 90% confidence interval for the standard deviation of birth weights at County General Hospital, if the standard deviation of the weights of the last 25 babies born there was 0.9 pounds and the birth weights at County General Hospital are normally distributed? Interpret your confidence interval.ANS: 90% implies α=0.1; consider the areas to the right of the critical values 1-α/2 = 1- 0.1/2 = 0.95 and α/2= 0.1/2 = 0.05; df = n – 1 = 25 – 1 = 24. χL2=13.848; χR2=36.41525-10.9236.415<σ<25-10.9213.848 0.73 < σ < 1.18We are 90% confident that the true population standard deviation is included in the interval between 0.73 and 1.18. ................
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