Section 11-2 8. Flat Tire and Missed Class. A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire that went flat? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the ability of the four students to select the same tire when they really didn’t have a flat?Left Front=11, Right Front=15, Left Rear= 8, Right Rear=618. American Idol. The contestants on the TV show “American Idol” try to win a singing contest. At one point, the web site WhatNotToSing.com listed the actual numbers of eliminations for different orders of singing, and the expected number of eliminations was also listed. The results are in the table below. Use a 0.05 significance level to test the claim that the actual eliminations agree with the expected numbers. Does there appear to be support for the claim that the leadoff singers appear to be at a disadvantage?Singing Order 1 2 3 4 5 6 7 thru 12Eliminations 20 12 9 8 6 5 9Expected 12.9 12.9 9.9 7.9 6.4 5.5 13.5 EliminationsSection 11-318. Baseball Player Births. In his book “Outliners”, author Malcolm Gladwell argues that more baseball players have birthdates in the months immediately following July 31, because that was the cutoff date for non-school baseball leagues. Here is a sample of frequency counts of months of birthdates of American-born major league baseball players starting with January: 387, 329, 366, 344, 336, 313, 313, 503, 421, 434, 398, and 371. Using a 0.05 significance level, is there sufficient evidence to warrant rejection of the claim that American born major league baseball players are born in different months with the same frequency? Do the sample values appear to support Gladwell’s claim?Section 12-2Popular Tree Weights. Weights (kg) of popular trees were obtained from trees planted in a sandy and dry region. The trees were given different treatments identified in a table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Use a 0.05 significance level to test the claim that the four treatment categories yield popular trees with the same mean weight. Is there a treatment that appears to be most effective in the sandy and dry region?No Treatment= 1.21, 0.57, 0.56, 0.13, 1.30Fertilizer= 0.94, 0.87, 0.46, 0.58, 1.03Irrigation= 0.07, 0.66, 0.10, 0.82, 0.94Fertilizer and irrigation= 0.85, 1.78, 1.47, 2.25, 1.64Section 8-228. Original Claim: Women have heights with a standard deviation equal to 5.00cm. The hypothesis test results in a P-value of 0.0055. (assume a significance level of =0.05)Section 8-36. Identify the indicated values or interpret the given display. Use the normal distribution as a approximation to binominal distribution. (assume a 0.05 significance level)Guns in the Home In a Gallup poll of 1003 randomly selected subjects, 373 said that they have a gun in their home. Info:Test of p = 0.35 vs p not = 0.35Variable – Guns X – 373N – 1003Sample p – 0.37188495% CI – (0.341974, 0.401795)Z-value – 1.45P – value – 0.146Section 8-420. Ages of Race Car Drivers. Listed below are the ages of randomly selected race car drivers. Use a .05 significance level to test the claim that the mean age of all race car drivers is greater than 30 years.32 32 33 33 41 29 38 32 33 23 27 45 52 29 25Section 8-512. Analysis of Pennies. In an analysis investigating the usefulness of pennies, the cents portions of 100 randomly selected checks are recorded. The sample has mean of 47.6 cents and a standard deviation of 35.5 cents. If the amounts from 0 cents to 99 cents are all equally likely, the mean is expected to be 49.5 cents and the population standard deviation is expected to be 28.866 cents. Use a 0.01 significance level to test the claim that the sample is from a population with a standard deviation equal to 28.866. If the amounts from 0 cents to 99 cents are all equally likely, is the requirement of a normal distribution satisfied? If not, how does that effect the conclusion?Section 4-4Among 143 subjects with positive test results, there are 24 false positive results: among 157 negative results, there are 3 false negative results.24. Screening for Marijuana Use. If 2 of the subjects are randomly selected without replacement, what is the probability that they both had correct test results (either true positive or true negative)? Is such an event unlikely?Section 4-530. Redundancy in Aircraft Radios. The FAA requires that commercial aircraft used for flying in instrument conditions must have two independent radios instead of one. Assume that for a typical flight, the probability of a radio failure is 0.0035. What is the probability that a particular flight will be safe with at least one working radio? Why does the usual rounding rule of three significant digits not work here? Is the probability high enough to ensure flight safety?Section 4-634. Mega Millions. AS of this writing, the Mega Millions lottery is run in 42 states. Winning the jackpot requires that you select the correct five numbers between 1
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