Proportion. Lecture 25 Sections Fri, Oct 10, Hampden-Sydney College. Sampling Distribution of a Sample. Proportion. Robb T.

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1 PDFs n = s Lecture 25 Sections Hampden-Sydney College Fri, Oct 10, 2008

2 Outline PDFs n = s PDFs n = 4 5 s 6 7

3 PDFs n = s The of the In our experiment, we collected a total of 100 samples, each of size 5. The distribution of sample proportions we observed was Sample Frequency

4 Homework Review PDFs n = s The of the To describe this distribution, we could describe its shape and give the mean and standard deviation. Keep in mind, we are working with a sample of values of. That is, a sample of samples. This is not the entire population of all samples of size 5.

5 Homework Review PDFs n = s The of the The shape is approximately normal. To find the mean and standard deviation, we may use the TI-83. Enter the distinct values of in list L 1. Enter the frequencies in list L 2. Use 1-Var Stats as bee, but enter both lists: 1-Var Stats L 1, L 2 We get the mean and the standard deviation.

6 Homework Review PDFs n = s The of the Theory says that the shape is approximately normal and that µ = p and σ = We calculate µ = 0.66 and σ = p(1 p). n p(1 p) (0.66)(0.34) = = n 5 Our experimental results were very close to what the theory predicts.

7 Distributions PDFs n = s Definition ( a Statistic) The sampling distribution of a statistic is the distribution of values of that statistic over all possible samples of a given size n from the population. We may sample with or without replacement. For large populations, the difference is insignificant. For our purposes, it will be easier to sample with replacement (allowing repetitions).

8 The Sample PDFs n = s The letter p represents the population proportion. The symbol ( p-hat ) represents the sample proportion. So is a random variable. The sampling distribution of is the probability distribution of all the possible values of.

9 Example PDFs n = Suppose that 2/3 of all males wash their hands after using a public restroom. Suppose that we take a sample of 1 male. Find the sampling distribution of. s

10 Example W W 1 2/3 2/3 PDFs n = s 1/3 N N 0 1/3

11 Example PDFs n = Let x be the number of males (in the sample) who wash. The probability distribution of x is x P(x) 0 1/3 = /3 = s

12 Example PDFs n = s Let be the sample proportion of males who wash. ( = x 1.) The sampling distribution of is P() 0 1/3 = /3 =

13 Example PDFs n = Now we take a sample of 2 males, sampling with replacement. Find the sampling distribution of. s

14 Example 2/3 W 2/3 1/3 W N WW 2 4/9 WN 1 2/9 PDFs n = s 1/3 N 2/3 W NW 1 2/9 1/3 N NN 0 1/9

15 Example PDFs n = s Let x be the number of males (in the sample) who wash. The probability distribution of x is x P(x) 0 1/9 = /9 = /9 =

16 Example PDFs n = s Let be the sample proportion of males who wash. ( = x 2.) The sampling distribution of is P() 0 1/9 = /2 4/9 = /9 =

17 Example PDFs n = Now we take a sample of 3 males, sampling with replacement. Find the sampling distribution of. s

18 Example 2/3 W WWW 3 8/27 2/3 W 2/3 1/3 W N 1/3 2/3 1/3 N W N WWN 2 4/27 WNW 2 4/27 WNN 1 2/27 PDFs n = s 1/3 N 2/3 1/3 W N 2/3 1/3 2/3 1/3 W N W N NWW NWN NNW NNN /27 2/27 2/27 1/27

19 Example PDFs n = s Let x be the number of males (in the sample) who wash. The probability distribution of x is x P(x) 0 1/27 = /27 = /27 = /27 =

20 Example PDFs n = s Let be the sample proportion of males who wash. ( = x 3.) The sampling distribution of is P() 0 1/27 = /3 6/27 = /3 12/27 = /27 =

21 Samples of Size n = 4 PDFs n = s If we sample 4 males, then the sample proportion of males who wash has the following distribution. P() / / /

22 Samples of Size n = 5 PDFs n = s If we sample 5 males, then the sample proportion of males who wash has the following distribution. P() / / / /

23 PDF n = 1 The pdf of when n = 1. PDFs n = s 0 1

24 PDF n = 2 The pdf of when n = 2. PDFs n = s 0 1/2 1

25 PDF n = 3 The pdf of when n = 3. PDFs n = s 0 1/3 2/3 1

26 PDF n = 4 The pdf of when n = 4. PDFs n = s 0 1/4 2/4 3/4 1

27 PDF n = 5 The pdf of when n = 5. PDFs n = s 0 1/5 2/5 3/5 4/5 1

28 PDF n = 6 The pdf of when n = 6. PDFs n = s 0 1/6 2/6 3/6 4/6 5/6 1

29 PDF n = 8 The pdf of when n = 8. PDFs n = s 0 1/8 2/8 3/8 4/8 5/8 6/8 7/8 1

30 PDF n = 10 The pdf of when n = 10. PDFs n = s 0 2/10 4/10 6/10 8/10 1

31 PDF n = 12 The pdf of when n = 12. PDFs n = s 0 2/12 4/12 6/12 8/12 10/12 1

32 PDF n = 15 The pdf of when n = 15. PDFs n = s 0 3/15 6/15 9/15 12/15 1

33 PDF n = 20 The pdf of when n = 20. PDFs n = s 0 4/20 8/20 12/20 16/20 1

34 PDF n = 30 The pdf of when n = 30. PDFs n = s 0 5/30 10/30 15/30 20/30 25/30 1

35 and Conclusions PDFs n = Observation The values of are clustered around p. Conclusion A randomly selected is probably close to p. s

36 and Conclusions PDFs n = s Observation As the sample size increases, the clustering becomes tighter. Conclusion Larger samples give better estimates. We can make the estimates of p as good as we want, provided we make the sample size large enough.

37 and Conclusions PDFs n = s Observation The distribution of appears to be approximately normal. Conclusion We can use the normal distribution to calculate just how close to p we can expect to be. However, we must know the values of µ and σ the distribution of. That is, we have to quantify the sampling distribution of.

38 s PDFs n = s Theorem ( s) For any population, the sampling distribution of has the following mean and standard deviation: µ = p p(1 p) σ =. n Furthermore, the sampling distribution of is approximately normal, provided n is large enough. n is large enough if np 5 and n(1 p) 5.

39 PDFs n = s Suppose that 54% of the voters support Barack Obama and 46% of them support John McCain, but we do not know that. We are trying to estimate the proportion of the voters who support John McCain. If we survey a random sample of 1000 people, how likely is it that our error will be no greater than 4%?

40 PDFs n = s In this case, is normal with mean µ = 0.46 and standard deviation (0.46)(0.54) σ = = Theree, P( ) is normalcdf(.42,.50,.46,.0158) =

41 Hypothesis Testing PDFs n = s Now suppose that we are trying to decide between the two hypotheses: H 0 : McCain has the support of 46% of the voters. H 1 : McCain has the support of 50% of the voters. Using a sample size of n = 100, draw the graphs of the sampling distributions. Calculate α and β.

42 Hypothesis Testing Now change the sample size to Draw the graphs and calculate α and β. PDFs n = s

43 PDFs n = Homework Read Sections , pages Exercises 7-14, page 526. s