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Text Sections: 4.
So far we have seen that, at least for some simple situations, we can develop mathematical tools to assess the likelihood that an event will occur. We now want to start pushing our techniques towards situations we are likely to encounter when we want to do surveys or polls.
For example, suppose you have heard that “left brain” versus “right brain” issues may be different for accountants as compared with artists. So, your research question might be to see what the proportion of “lefties” versus “righties” is like for accountants. How do we think about this situation?
The model we will use is based upon a coin toss with a biased coin. Since there are many accountants in America, walking up to one of them and asking “Are you a leftie?” is a lot like tossing a coin and seeing whether it comes up HEADS, especially if we ask these accountants randomly in a simple random sample (remember this idea from the first chapter. We think of this like having all the accountants in America write their names on pieces of paper and then putting these names in a big hat and choosing, say 10 of them).
Some questions come up immediately:
- Since we cannot, in a sample , be sure to get the correct proportion (i.e. the sample proportion will not generally match the population proportion ) how close do we need to be?
- Once we know that we need to be within, say, 3% of the true proportion of accountant lefties, how many people do we need in our sample? (The more the better, of course, but after awhile this gets expensive and time consuming).
- Since we cannot be sure that we’ll be within 3% of the true number, how confident can we be that our estimate will be within 3% of the true number?
So, how to proceed? Let’s first try to understand coin tosses (which are easy to think about and mathematically “clean”), then come back to samples.
Think about a simple case where we toss a fair coin 4 times. Recall our sample space:
HHHH HTHH THHH TTHH
HHHT HTHT THHT TTHT
HHTH HTTH THTH TTTH
HHTT HTTT THTT TTTT
Since our coin is fair, we can find the probability of obtaining 0, 1, 2, 3, 4, or 5 HEADS just by counting. Use the Classical Notion of Probability to say that, since there are 16=2^4 possible outcomes, the probability that in our sample we will have, say 3 HEADS is
We can do this for each of these to obtain
Number of HEADS
probability 1/16 4/16 6/16 4/16 1/
If you are very patient you can repeat this procedure for a sample of size but after that it gets a little rough. What can we do for a sample of size 10? Of size 500? Here’s the first idea we can use.
Conditional Probability Very often we would like to compute the probability of an
event in the presence of knowledge about some other, perhaps related, event. For example, suppose that you ask individuals, at random, in the Mall, whether or not they own a snowmobile. Do you expect to get the same frequency of YES responses if:
- You are in the Mohawk Valley? (Lots of snow, plenty of space.)
- You are in the center of Manhattan? (Much less snow, much less space.)
To see how we might assign probabilities conditionally consider the following example. One of our students, Mary P., did a fine job collecting blood pressure and temperature data for men and for women in a sample at her workplace (Nice Job, MP!) If we consider just gender and whether the systolic blood pressure is 120 or above versus less than 120 we get the following table:
MEN WOMEN
120 or ABOVE 5 8 13
BELOW 120 5 2 7
10 10 20
Suppose you randomly select an individual from this sample. What is the probability that their systolic BP is 120 or more?
Since there are 20 people in the class, and since the number of individuals with 120 or more BP is 13, this probability is.
Example : Suppose you are going to take a bus trip from Utica to Buffalo. The bus
company has furnished you with the following data:
prob(Leave On Time) = prob(LOT) = 0.
prob(Arrive On Time) = prob(AOT) = 0. prob(Leave on Time and Arrive On Time) = 0.
Calculate the following:
- Your bus has just left on time. What is the probability that it will arrive on time? Since our bus has just left on time we will condition on that probability. To get the probability that it will arrive on time given that we leave on time we set up our formula just like before:
It seems like your chances of arriving on time are enhanced if you leave on time.
- Your bus has just left late. What is the probability that it will arrive on time?
This is a little trickier, so it’s best to think in terms of a picture here. Remember that we can draw a Venn diagram to help us.
Region
Leave on Time Arrive on Time
Region Region4 Region
Region 1 represents neither Arriving on Time nor Leaving on Time Region 2 represents not Arriving on Time but Leaving on Time
Region 3 represents Arriving on Time but not Leaving on Time Region 4 represents Arriving on Time and Leaving on Time
What are the sizes of these regions? We have been told that the event “Leave on Time” has a “size” or probability of 0.80. That is, the total size of regions 2 and 4 together is 0.80. We also know that the size of region 4 “Leave on Time and Arrive On Time” is 0.65. Since regions 2 and 4 do not overlap we must have that the size of Region 2 is 0.80-0.65 = 0.15. So, the probability that we Leave on Time but do not Arrive on Time is 0.15.
Work the same way to get the probability of Region 3: “Arriving on Time but not Leaving on Time”. – Region 4 together with Region 3 make up the event Arrive on Time. So,
Looking back on our numbers this means that
And so
How large is Region 1? We know that the 4 non-overalpping regions have sizes that must add up to 1 (the size of our sample space) so
It is not very likely to leave late and also to arrive late.
Back to our question. Your bus has just left late. What is the probability that it will
arrive on time? We can now see that
