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Chapter 9.2: Sampling Proportions
Chapter 9.2 Homework Name_____________________
Done on Time:____/5 Completed:____/5 1. A USA Today poll asked a random sample of 1012 U.S. adults what they do with milk in the bowl after they have eaten the cereal. Of the respondents, 67% said that they drink it. Suppose that 70% of U.S. adults actually drink the cereal milk. (a) Find the mean and standard deviation of the proportion p of the sample who say they drink the cereal milk.
The mean is μp=p=0.7 and the standard deviation is σp=p(1-p)n=0.7×0.31012=0.0144. (b) Explain why you can use the formula for standard deviation of p in this setting (Rule of Thumb 1) The population (all U.S. adults) is clearly at least 10 times as large as the sample (the 1012 surveyed adults). (c) Check that you can use the Normal approximation for the distribution of p (Rule of Thumb 2)
The two conditions, np = 1012×0.7 = 708.4 > 10 and n(1 − p) = 1012×0.3 = 303.6 > 10, are both satisfied.

(d) Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say they drink the cereal milk. Do you have any doubts about the results of this poll?

P(p≤0.67) = P(Z ≤ −2.08) = 0.0188. This is a fairly unusual result if 70% of the population actually drinks the cereal milk.

(e) What sample size would be required to reduce the standard deviation of the sample proportion to one-half the value you found in (a)?
To half the standard deviation of the sample proportion, multiply the sample size by 4; we would need to sample 1012×4 = 4048 adults. (f) If the pollsters had surveyed 1012 teenagers instead of adults, do you think the sample proportion p would have been greater than, equal to, or less than .67? Explain.It would probably be higher, since teenagers (and children in general) have a greater tendency to drink the cereal milk. 2. The Gallup Poll

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