Stat 11
February 17, 2006
Homework #4 - SOLUTIONS
(Jargon from section 3.2 on experiments)
3.12
a.
Subjects – the 22000 male physicians.
Factor – taking aspirin (or more precisely, being assigned to take aspirin)
Levels – only two; taking aspirin or not
Response Variable – whether a subject had a heart attack during the observation period
b.
Group 1 Treatment 1
11000 physicians Aspirin twice/day
Random Compare
allocation Group 2 Treatment 2 Heart Attack
11000 physicians Placebo Frequency
c.
It means that the difference between the groups probably didn’t arise from randomness. (It doesn’t say, beyond that, how big the difference was or whether it may have been due to some non-obvious bias in the experiment.)
3.26 (statistical significance is defined on page 203; much more will be said about it)
In fact, the actual difference in Monday market returns (if it exists) is too small to see without a careful study, and too small to do anything about. The study is saying that they did see a difference between 1st-2nd-3rd Mondays and 4th-5th Mondays, stock market returns were in fact higher on the first group, and by enough that it isn’t just due to randomness. The difference between 1st-2nd-3rd Mondays and zero returns, however, was small enough that it might have been due to just randomness.
(Kinds of samples)
3.36 For this problem, identify all four: POPULATION, SAMPLING FRAME, SAMPLE AS SELECTED, and ACTUAL SAMPLE. They’re all different. Also, say what the non-response rate is. (The text defines sampling frames in problem 3.55.)
population: all local business
sampling frame: the business listed in the telephone book
sample (as selected): 150 business drawn at random from the sampling frame
actual sample: 73 business who returned completed questionnaires
(That’s a non-response rate of 77/150, or about 51%.)
3.48 (say why this sample fails to satisfy the definition of SRS)
There are 45 groups of 200 students who might be chosen: Once the first student is chosen (from #’s 1-45) the entire sample is determined. Each group has a 1/45 chance. Each student is in one and only one group, so each student has a 1/45 chance of being in the sample.
But in an SRS, each group of any 200 students would have a chance to be chosen. There are zillions of possible groups, and many (such as, the first 200 students, #’s 1-200) can’t be picked by this method.
3.52 (say why this sample fails to satisfy the definition of SRS)
Every partygoer has the same chance, but not every group of 5 has a chance --- for example, the group that consists of the 5 oldest students can’t be picked by this method.
This sample is better than an SRS (for most purposes). But, it isn’t an SRS.
(Bias and variability)
3.66
a. large negative bias, high variability (by the scale of these problems)
b. not much bias, low variability
c. not much bias, high variability
d. large positive bias, low variability
(A sampling distribution)
3.73 This problem uses the “probability” applet at the text’s website. (You could use the Excel worksheet from class, too, if you dare.) Set up the true probability (0.6), the sample size (25) and start pushing buttons. This would be a good problem to do in a group, but please draw your own histogram.
Your histograms should be roughly normal (but with lots of random irregularity, even with 50 trials each).
The first should be centered near 15 (because that’s 0.6 times 25)
and the second near 6 (0.2 times 25).
That shows that ON AVERAGE, samples of size 25 will give you the right mean, but that individual samples might be off by quite a bit.
(Probability)
4.7 Pascal and Fermat really did work on this; it was harder for them than it is for you. You may use the applet if you like, OR you may construct any rational argument for why the probability of seeing a “6” is more than 50%, less than 50%, or exactly 50%.
It’s more than 50%, but it would take you a lot of trials to be sure of that by experiment.
Here’s an argument: the chance of getting a NON-6 on one die is 5/6. The chance of getting NON-6’s on all four dice is (5/6) times (5/6) times (5/6) times (5/6), or 625/1296, which is less than 50%. The chance of getting at least one 6 is the complement of that event, so its probability is more than 50%.
4.13 (complement rule, probability of an “event”)
0.04 (or, “4 percent”). (That’s what it takes to make to make the four outcome probabilities add to 1)
4.25 (essential knowledge for every probability student)
a. 1/38 for any one slot (ideally)
b. 18/38, because that event includes 18 of the outcomes (=slots)
c. 12/38, because that event includes 12 of the outcomes.
(end)