1. Sampling Distributions of Proportions

2. Toss a penny 20 times and record the number of heads. Calculate the proportion of heads & mark it on the dot plot on the board. What shape do you think the dot plot will have? The dotplot is a partial graph of the sampling distribution of all sample proportions of sample size 20. If I found all the possible sample proportions – this would be approximately normal!

3. Sampling Distribution Is the distribution of possible values of a statistic from all possible samples of the same size from the same population In the case of the pennies, it’s the distribution of all possible sample proportions (p) We will use: p for the population proportion and p-hat for the sample proportion

4. Suppose we have a population of six people: Melissa, Jake, Charles, Kelly, Mike, & Brian What is the proportion of females? What is the parameter of interest in this population? Draw samples of two from this population. How many different samples are possible? Proportion of females 6 C 2 =15 1/3

5. Find the 15 different samples that are possible & find the sample proportion of the number of females in each sample. Alice & Ben.5 Alice & Charles.5 Alice & Denise 1 Alice & Edward.5 Alice & Frank.5 Ben & Charles 0 Ben & Denise.5 Ben & Edward 0 Ben & Frank 0 Charles & Denise.5 Charles & Edward 0 Charles & Frank 0 Denise & Edward.5 Denise & Frank.5 Edward & Frank 0 Find the mean & standard deviation of all p-hats. How does the mean of the sampling distribution (  p-hat ) compare to the population parameter (p)?  p-hat = p

6. Formulas: The mean of the sampling distribution. The standard deviation of the sampling distribution.

7. Formulas: These are found on the formula chart!

8. Does the standard deviation of the sampling distribution equal the equation? NO - WHY? We are sampling more than 10% of our population! If we use the correction factor, we will see that we are correct. Correction factor – multiply by So – in order to calculate the standard deviation of the sampling distribution, we MUST be sure that our sample size is less than 10% of the population!

9. Assumptions (Rules of Thumb) Use this formula for standard deviation when the population is sufficiently large, at least 10 times as large as the sample. Sample size must be large enough to insure a normal approximation can be used. We can use the normal approximation when np > 10 & n (1 – p) > 10

10. Assumptions (Rules of Thumb) Sample size must be less than 10% of the population (independence) Sample size must be large enough to insure a normal approximation can be used. np > 10 & n (1 – p) > 10

11. Why does the second assumption insure an approximate normal distribution? Suppose n = 10 & p = 0.1 (probability of a success), a histogram of this distribution is strongly skewed right! Remember back to binomial distributions Now use n = 100 & p = 0.1 (Now np > 10!) While the histogram is still strongly skewed right – look what happens to the tail! np > 10 & n(1-p) > 10 insures that the sample size is large enough to have a normal approximation!

12. Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time. The bank recently approved 200 loans. What are the mean and standard deviation of the proportion of clients in this group who may not make payments on time? Are assumptions met? What is the probability that over 10% of these clients will not make payments on time? Yes – np = 200(.07) = 14 n(1 - p) = 200(.93) = 186 1- (Normcdf(-∞, 0.1, 0.07, 0.01804) OR: Ncdf(.10, 1E99,.07,.01804) =.0482

13. Suppose one student tossed a coin 200 times and found only 42% heads. Do you believe that this is likely to happen? No – since there is approximately a 1% chance of this happening, I do not believe the student did this. np = 200(.5) = 100 & n(1-p) = 200(.5) = 100 Since both > 10, I can use a normal curve! Find  &  using the formulas.

14. Assume that 30% of the students at HH wear contacts. In a sample of 100 students, what is the probability that more than 35% of them wear contacts? Check assumptions!  p-hat =.3 &  p-hat =.045826 np = 100(.3) = 30 & n(1-p) =100(.7) = 70 P(p-hat >.35)= Ncdf(.35, 1E99,.3,.045826) =.1376 Example #3 Example #1

15. STATE: We want to know the probability that a random sample yields a result within 2 percentage points of the true proportion. We want to determine

16. Example #1 PLAN: We have drawn an SRS of size 1500 from the population of interest. The mean of the sampling distribution of p-hat is 0.35:

17. Example #1 PLAN: We can assume that the population of first-year college students is over 15,000, and are safe to use the standard deviation formula: In order to use a normal approximation for the sampling distribution, the expected number of successes and failures must be sufficiently large: Therefore,

18. Example #1 DO: Perform a normal distribution calculation to find the desired probability:

19. Example #1 CONCLUDE: About 90% of all SRS’s of size 1500 will give a result within 2 percentage points of true proportion.