1. Sampling Distributions Adapted from Exploring Statistics with the TI-83 by Gail Burrill, Patrick Hopfensperger, Mike Koehler

2. A company wants to know what percent of consumers like the ads for their product. A TV network would like information on the proportion of the TV viewing public that watches their programming. In both cases, information is needed from a large number people, but it would not be possible to ask the entire populations for their opinions. Samples may be made and the information from the samples may be used to make inferences about the populations.

3. Suppose we are interested in consumers opinions of advertisements that you are showing on TV that talk about a new shampoo you want to market. We can try to talk to all the consumers who have watched the ads but that would be too big a task. Instead, we will ask, say, a sample of 20. Okay, go out right now and ask 20 people who have seen the ads whether or not they like it. Well, maybe you can’t do that right now because you have to stay in class! So, instead of actually asking the people we can simulate what sample responses would be. To do this, we take into account some previously known percentage. Assume for this case that 40% like the ads. Let’s make a note here that you will always need to know some percentage and it will be given in the problem sets. Since this scenario is binomial (check the conditions you learned in Module 8) we will use a random binomial generator. The next slide shows how……..

4. Remember, we are interested in the percent who did like the ads and we know that in the past 40% have liked our ads. We will use the TI-83/84 command randBin to simulate a sample of 20 persons, 40% of whom liked the ads. (RandBin is found under the menu.) So, in this sample, 9 out of 20 persons liked the ads. That’s 45% - a little more than 40%. If we did this many times and recorded each percent (or count) that liked the ad we could form a distribution of that. That’s known as a Sampling Distribution Notice that the command is “randBin(# in sample, percent-in decimal form, #trials to run)” The last part “# trials to run” is only input when there is going to be more than one trial run. See the next slide for that.

5. What if we did this many times over and recorded the number who liked the ads each time. Let’s run the simulation for 100 times! An easy way to do this is to specify 100 runs and save the results to L 1. Notice the commands – randBin(20,.4.100) = randBin(# in sample, percent-in decimal form, # trials to run) Make a histogram of the results and calculate the mean and standard deviation. The mean is 8.09 and the sample standard deviation is 2.374.

6. Now, lets find the 5 th and the 95 th percentiles. For this set it is easy since we are running 100 trials. Recall that a percentile represents a place in a list. The 5 th percentile is the number in the list in which 5% of all the numbers are that value or lower. I just ran this in my calculator and got “5” as my 5 th percentile. Since there are 100 numbers in my L 1 then the fifth one is my 5 th percentile. Yours may be slightly different because each calculator runs the simulation uniquely. My 95 th percentile is the 95 th number in that list. For me on my run I got 12. Let’s also keep in mind what these numbers represent. Each of our 100 numbers in L 1 represents the number of people out of 20 who like our ads. So, I can say that in general if I were to ask 20 people then 5% of the time I would get 5 or fewer “yes” answers and 95% of the time I would get 12 or fewer “yes” answers. I can also say, then, that 90% of the time I will get “yes” answers between 5 and 12 – if I asked 20 people each time.

7. So you may have a couple of questions: 1.Why did I use “randBin”. The answer is that this is a binomial scenario and since we want to make a distribution of “yes” answers out of 20 then we make a binomial distribution. If the scenario was not binomial we would use a normal distribution “normalcdf” 2.Why 5 th percentile and 95 th percentile? Why not 10 th and 90 Th or some other combination. The answer to this is that it is common to look at the middle 90% of all distributions. This is a concept we will learn more about in Module 10 so for now, just learn how to crunch the numbers.

8. Now you will have a chance to try a few problems in class.

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