Part 5 Review solutions

# Part 5 Review solutions - Review of Part V 379 Review of...

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Review of Part V 379 Review of Part V – From the Data at Hand to the World at Large 1. Herbal cancer. H 0 : The cancer rate for those taking the herb is the same as the cancer rate for those not taking the herb. pp Herb Not Herb Not =− = () or 0 H A : The cancer rate for those taking the herb is higher than the cancer rate for those not taking the herb. Herb Not Herb Not >− > or 0 2. Colorblind. a) 10% condition: The 325 male students are probably representative of all males, and 325 students are less than 10% of the population of males. Success/Failure condition: np = (325)(0.08) = 26 and nq = (325)(0.92) = 299 are both greater than 10, so the sample is large enough. Since the conditions have been satisfied, a Normal model can be used to model the sampling distribution of the proportion of colorblind men among 325 students. b) µ ˆ p == p 0.08 σ ( ˆ) p pq n (. ) . 008 092 325 0 015 c) d) According to the Normal model, we expect about 68% of classes with 325 males to have between 6.5% and 9.5% colorblind males. We expect about 95% of such classes to have between 5% and 11% colorblind males. About 99.7% of such classes are expected to have between 3.5% and 12.5% colorblind males. 3. Birth days. a) If births are distributed uniformly across all days, we expect the number of births on each day to be np = () ≈ 72 10 29 1 7 . . b) 10% condition: The 72 births are likely to be representative of all births at the hospital with regards to day of birth, and 72 births are less than 10% of the births. Success/Failure condition: The expected number of births on a particular day of the week is np = 72 10 29 1 7 . and the expected number of births not on that particular day is nq = 72 61 71 6 7 . . These are both greater than 10, so the sample is large enough.

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380 Part V From the Data at Hand to the World at Large Since the conditions have been satisfied, a Normal model can be used to model the sampling distribution of the proportion of 72 births that occur on a given day of the week. µ ˆ . p p ==≈ 1 7 0 1429 σ ( ˆ) p == pq n () . 1 7 6 7 72 0 04124 There were 7 births on Mondays, so ˆ . p =≈ 7 72 0 09722. This is only about a 1.11 standard deviations below the expected proportion, so there’s no evidence that this is an unusual occurrence. c) The 17 births on Tuesdays represent an unusual occurrence. For Tuesdays, ˆ . p 17 72 0 2361, which is about 2.26 standard deviations above the expected proportion of births. There is evidence to suggest that the proportion of births on Tuesdays is higher than expected, if births are distributed uniformly across days. d) Some births are scheduled for the convenience of the doctor and/or the mother. 4. Polling. a) No, the number of votes would not always be the same. We expect a certain amount of variability when sampling. b) This is NOT a problem about confidence intervals. We already know the true proportion of voters who voted for Clinton. This problem deals with the sampling distribution of that proportion.
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## This note was uploaded on 05/20/2008 for the course STAT 50 taught by Professor Weinstein during the Spring '08 term at Harvard.

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Part 5 Review solutions - Review of Part V 379 Review of...

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