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Material Type: Exam; Class: Statistical Methods; Subject: Statistics; University: Virginia Polytechnic Institute And State University; Term: Fall 2011;
Typology: Exams
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Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) Part I: Definition ( Note: Put your answer above the line ) (50 pts)
b. If we are interested in measuring peopleโs addiction to watching TV, which is denoted by TVADDIC. What is in below shows how we get TVADDIC from TVWATCH.
๐๐๐ด๐ท๐ท๐ผ๐ถ = ๏ฟฝ
Then, TVADDIC is B_ A. Continuous Random Variable B. Discrete Random Variable
Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) F. It is binary.
Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) Part II: Application ( Note: Keep numbers to two decimal places ) (50 pts)
b. (4 pts) At least how many days should the longest 25% of all pregnancies last? We need to compute the 75th^ percentile of Y, i.e., ๐.๐๐ ๐.๐๐ = ๐.๐๐^ ๐๐โ ๐๐๐ = ๐. ๐๐ โ ๐.๐๐ = ๐. ๐๐ โ ๐๐ + ๐๐๐ = ๐๐๐. ๐๐ Therefore, at least 276.88 days should the longest 25% of all pregnancies last.
c. (3 pts) Suppose a certain obstetrician is currently providing prenatal care to 60 pregnant women. Let ๐ฆ๏ฟฝ represent the mean length of their pregnancies. What is the distribution of ๐ฆ๏ฟฝ? Specify the distribution, mean, and standard error. Y is exactly normal โน ๐๏ฟฝ^ is exactly normal ๐ฌ(๐๏ฟฝ) = ๐ฌ(๐) = ๐๐๐ ๐บ๐ฌ(๐๏ฟฝ) = ๐บ๐ซ(๐) โ๐^
Therefore, ๐๏ฟฝ~๐ต(๐๐๐, ๐๐^
๐ ๐๐ =^ ๐.^ ๐๐) d. (4 pts) What is the probability that the mean duration of the 60 patientsโ pregnancies in part (c) will be between 260 and 270 days? From part (c), we know that ๐๏ฟฝ~๐ต(๐๐๐, ๐๐ ๐/๐๐)
Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) Therefore, ๐ท(๐๐๐ โค ๐๏ฟฝ โค ๐๐๐) = ๐ท ๏ฟฝ๐๐๐ โ ๐๐๐๐๐ โ๐๐โ โค ๐
e. (5 pts) How does your answer in part (d) compare to your answer in part (a)? Does this relationship make sense? Why or why not? The answer in part (d) is much larger than that in part (a). It makes sense in that the mean is less variable than an individual value, so for a fixed range (260 days to 270 days in this case) we would expect the mean to be in the region with high probability.
f. (5 pts) The duration of human pregnancies may not always follow a Normal distribution. Suppose that we made a mistake and the correct distribution for human pregnancies is in fact skewed, does that change your answers in (a) and (d)? Briefly explain why or why not for each. For part (a), the answer changes because it cannot be computed via using the z-table without knowing the population distribution is normal. For part (d), the answer does not change because of the central limit theorem. Since we know the sample size is 60 (>=30), the sampling distribution is approximately normal even though the population distribution is not known.
Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) (1-0.8311, 1-0.7329)=(0.1689, 0.2761) โ (0.17, 0.28) e. (15 pts) Suppose the true proportion that supports Road Runner is 70%. Specify the Population distribution , Sampling distribution , and Data distribution. (Hint: Let X=0 to denote the person votes for Wile E. Coyote, and let X=1 to denote the person votes for Road Runner) Population Distribution X 0 1 P(x) 0.3 0.
Data Distribution X 0 1 P(x) 0.2225 0.
Sampling Distribution ๐๏ฟฝ~๐ต(๐. ๐, ๐๐๐๐.^ ๐๐ = ๐. ๐๐๐๐๐๐)
Tuesday, Nov 8, 2011 (9:30AM โ 10:45AM) Extra Credit
๐๐๐ธ = ๐ถ๐ โ ๐๐ธ = ๐1โ๐ผ 2 โ ๏ฟฝ๐ฬ๐๐๏ฟฝ
To guarantee a margin of error of at most m, We set ๐๐๐ธ โค ๐
โ ๐1โ๐ผ 2 โ ๏ฟฝ๐ฬ๐๐ ๏ฟฝโค ๐
โ ๐ โฅ ๐1โ๐ผ/^2