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Dive into diverse statistical scenarios including criminality, cigarette filters, sports performance, and more to understand p-values, distributions, means, and standard deviations in sample data. Enhance your statistical skills with practical examples.
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AP Statistics Sample Means and Proportions Scavenger Hunt Review
0.0423 • A study of chromosome abnormalities and criminality. Of the 4096 men with normal chromosomes, 381 had criminal records, while 8 of the 28 men with abnormal chromosomes had criminal records. Some experts believe chromosome abnormalities are associated with increased criminality. Find the p-value.
.0002538 • The developer of a new filter for filter-tipped cigarettes claims that it leaves less nicotine in the smoke than does the current filter. Because cigarette brands differ in a number of ways, he tests each filter on one cigarette of each of nine brands and records the difference in nicotine content (current filter – new filter). From the nine cigarettes they get a mean difference of1.32 mg and a standard deviation of 2.35 mg. Find the p-value.
0.065229 • LeRoy, a starting player for a major college basketball team, made only 40% of his free throws last season. In the first eight games of this season, LeRoy made 25 free throws in 40 attempts. You want to investigate whether LeRoy’s work over the summer will result in a higher proportion of free-throw successes this season. What is the p-value?
0.001838 • The average age of cars owned by residents of a small city is 4 years with a standard deviation of 2.2 years. A SRS of 400 cars is selected, and the sample mean age computed. Describe the distribution, the mean and standard deviation of the sample mean.
Approx normal with mean = 4, std dev = 0.11 • A SRS of 100 Americans found that 61% were satisfied with service provided by the dealer from which they bought their car. Describe the distribution, mean, and standard deviation of the sample.
Approx normal withmean = 0.61, std dev = 0.049 • The incomes in a certain large population of college teachers have a normal distribution with mean $35,000 and standard deviation $5000. Four teachers are selected at random from this population to serve on a salary review committee. What is the probability that their average salary exceeds $40,000?
0.0228 • The strength of paper coming from a manufacturing plant is known to be 25 pounds/square inch with a standard deviation of 2.3. In a random sample of 40 pieces of paper, what is the probability that the mean strength is between 24.5 and 25.5 pounds/square inch?
0.8351 • A promoter knows that 23% of males enjoy watching boxing matches. In a random sample of 125 men, what is the distribution, mean and standard deviation of the sample?
Approx normal withmean = 0.23, std dev = 0.038 • A promoter knows that only 12% of females enjoy watching boxing matches. In a random sample of 125 women, what is the probability that more than 10% of the females enjoy watching boxing matches?
0.7549 • The average number of missed school days for students in public schools is 8.5 with a standard deviation of 4.1. In a sample of 200 public school students, what is the probability that the average number of days missed is less than 8 days?
Answers • Adult women & time for themselves: • Approx normal, μ= .47, σ = .016 • Np = 1025*.47 = 481.75 • Nq = 1025 *.53 = 543.25 • Internet access fees • Approx normal with μ= 28, σ = 0.447 • P(x-bar > 29) = 0.0126
Answers • Mail order company • Approx normal with μ= .9, σ = .03 • Np = .9 * 100 = 90, nq = .1 * 100 = 10 • P(p-hat ≤ .86) = 0.0912 • Age of cars • Approx normal, mean = 4, std dev = 0.11
Answers • Car Dealership service • Approx normal with mean = 0.61, std dev = 0.049 • Np = 61, nq = 39 • College teacher incomes • P(x-bar ≥ 40,000) = 0.228 with std dev = 2500
Answers • Paper strength • Std dev = .36 • P(24.5 ≤ x – bar ≤ 25.5) = .8351 • Males & boxing matches • Approx normal with mean = .23, std dev = 0.38 • Np = .23 * 125 = 28.75 • Nq = .77 * 125 = 96.25
Answers • Females & boxing matches • Approx normal with mean = .12, std dev = .029 • Np = 125 * .12 = 15 • Nq = 125 *.88 = 110 • P(p=hat ≥ 0.1) = .7548 • Missed school days • Std dev = .29 • P(x-bar < 8) = .0423