Stat 211 – T1, Fall 2012Name __________Key____________________Please read all questions carefully. Follow directions, present your answer legibly and show support workwhenever practical. Relax, pace yourself and be organized of thought and deed.1. For the data set {2, 11, 18, 19, 19, 21, 21, 22, 23, 23}, please determine the following:a. (3) The median: ̃b. (6) The cutoff point for extreme outliers:( )( )2. (3) State the formula for calculating a sample standard deviation? Please state the proper notation for labeling thesample standard deviation.√ ∑ ( ̅)3. (8) A medical testing laboratory performs urine tests to detect illegal drug use among ex-convicts enrolled in acourt-mandated drug-avoidance program. Laboratory statistics show that 90% of urine samples containing evidenceof illegal-drug use will test positive, and 98% of urine samples containing no evidence of illegal drug will use testnegative. Overall, about 35% of the ex-convicts in the program have evidence of illegal-drug use in their urine atany given time. Given that a urine sample tests positive, what is the probability that the sample actually containsevidence of illegal drug use?̅( | ) ( | ) ( | ̅) ( | ̅) ( ) ( ̅)Want ( | ), which is( | )( | ) ( )( | ) ( ) ( | ̅) ( ̅)( | )( )(( )( ) ( )( )

4. (7) Ninety nine percent of Americans have an office phone number or a personal phone number. If ninety percentof Americans have a personal phone number and fifty five percent have an office phone number, what percent of thepopulation has both a personal phone number and an office phone number? Are the events mutually exclusive?( ) ( ) ( ) ( )( )5. Among female Vermonters, blood cholesterol level is normally distributed with a mean of 195 mg/dl and astandard deviation of 14 mg/dl.a. (7) What is the probability that a randomly-selected female Vermonter will have a blood cholesterol levelbetween 180 and 205 mg/dl?( ) ( ) = normalcdf( )b. (7) What is the probability that a randomly-selected female Vermonter will have a blood cholesterol levelless than 185 mg/dl or greater than 190 mg/dl?( ) ( ) = normalcdf( )c. (7) In a random sample of 16 female Vermonters, what is the probability that the sample mean will exceed200 mg/dl?( ̅) (√) = normalcdf( )6. (7) United Airlines measured the flying time from Albuquerque to Denver for flight 448 on 10 randomly selecteddays. The mean flying time was 55 minutes and the sample standard deviation was 4 minutes. Compute the 90%confidence interval for the true mean flying time. You may assume that the flying time is normally distributed.√√( )7. (2) In general, if a confidence level is increased, the margin of error (moe) INCREASES8. (2) In general, as sample size increases, the margin of error (moe) DECREASES

9. (5) When reading an article, you encounter the line, “”The 90% confidence interval for the average temperature ofbasalt lava is (1,995, 2,245) degrees Fahrenheit. How do you interpret this line?We are 90% confident that the average temperature of basalt lava is between 1995 and 2245 degreesFarenheit.10. (8) The Vermont Department of Agriculture is planning a study of lead concentration in maple syrup produced in2009. Typically, the concentration of lead ranges from 50 parts per billion (ppb) to 450 ppb and is normallydistributed. The regulators would like to estimate the true mean lead concentration using a 95% confidence intervalwith a margin of error of 20 ppb. What sample size should they use?Therefore√( )( )Thusn = 9711. (8) Arsenic in drinking water is a major health concern. The current EPA standard of at most 50 ppb (parts perbillion) is considered by some public health researchers to be too high. In a particular community, the maximum isset at 15 ppb. To ensure compliance, the water is tested regularly. In a recent study, data from 12 random watersamples yielded a mean of 17 ppb and a standard deviation of 2 ppb. Is there statistical evidence that the arsenicconcentration exceeds the community standard? Perform the appropriate statistical test using α = 0.01. You mayassume that arsenic concentration has a normal distribution.H 0 : µ = 15H 1 : µ > 15Upper tail testα = 0.01√( )Reject H 0 in favor of H 1 at the one percent level of significance.12. (5) Given your conclusion in question 9, describe the hypothesis-testing error(s) that might have occurred, andlist one or two possible ramifications.Since we rejected the null hypothesis, the only mistake we could have made would be a type I or α error,which is erroneously rejecting a true null. The ramifications in this case would be that we would conclude thatthe drinking water has excessive amounts of arsenic (by the community’s standards) which might motivatethem to spend money to mitigate a problem which doesn’t actually exist.13. (3) What is power?Power is the ability to detect and reject a false null hypothesis. It is equal to 1-β, which could be seen as theprobability of not making a type II error.

14. (8) An epidemiologist is interested in studying the effects of cigarette smoking on serum cholesterol and believesthat smoking is associated with an increase in cholesterol. The epidemiologist wishes to test the null hypothesis H 0 :μ ≤ 180 mg/dl versus the alternative H a : μ > 180 mg/dl in a study of 150 smokers using α = 0.05. Suppose that thepopulation standard deviation is σ = 40 mg/dl. Compute the power of the test when μ a = 190 mg/dl.Zα = 1.645, therefore the ̅ value at the boundary of the rejection region is√(√)Power = 0.921715. (2) In the scenario in problem 12, if α is increased, β DECREASES16. (2) In the scenario in problem 12, if μ a is increased, β DECREASES

More magazines by this user
Similar magazines