Lecture 26 Section 8.4. Wed, Oct 14, 2009

Transcription

1 PDFs n = Lecture 26 Section 8.4 Hampden-Sydney College Wed, Oct 14, 2009

2 Outline PDFs n = 1 2 PDFs n =

3 Outline PDFs n = 1 2 PDFs n =

4 PDFs n = Exercise 8.12, page 528. Suppose that 60% of all students at a large university access course inmation using the Internet. (a) Sketch a picture of the distribution the possible sample proportions you could get based on a simple random sample of 100 students. (b) Use the rule normal distributions to complete the following statements: (i) There is a 68% chance that the sample proportion is between and. (ii) There is a 95% chance that the sample proportion is between and. (iii) It is almost certain that the sample proportion is between and.

5 PDFs n = Exercise 8.12, page 528. (c) Would it be likely to observe a sample proportion of 0.50, based on a simple random sample of size 100, if the population proportion were 0.60? Explain. (d) Sketch a picture of the distribution the possible sample proportions you could get based on a simple random sample of 400 students. (i) How does this picture differ from the one in part (a)? (ii) How will the increased sample size affect the range of values you gave in (i) (iii) of part (b)

6 Solution (a) For n = 100 students, the sketch is PDFs n =

7 PDFs n = Solution (b) First, compute µˆp and σˆp. µˆp = p = p(1 p) (0.60)(0.40) σˆp = = = n 100 (i) There is a 68% chance that the sample proportion is between = and = (ii) There is a 95% chance that the sample proportion is between (0.0490) = and (0.0490) = (iii) It is almost certain that the sample proportion is between (0.0490) = and (0.0490) =

8 PDFs n = Exercise 8.12, page 528. (c) The question should ask how likely it is to observe a sample proportion at least as low as The probability is P (ˆp 0.50), which is normalcdf(-e99,0.50,0.60,0.490) =

9 Exercise 8.12, page 528. (d) For n = 400 students, the sketch of ˆp is PDFs n =

10 PDFs n = Exercise 8.12, page 528. (d) (i) This distribution is only half as wide (and twice as tall). (ii) The standard deviation of ˆp is only half as much, so the answers are (i) There is a 68% chance that the sample proportion is between = and = (ii) There is a 95% chance that the sample proportion is between (0.0245) = and (0.0245) = (iii) It is almost certain that the sample proportion is between (0.0245) = and (0.0245) =

11 Outline PDFs n = 1 2 PDFs n =

12 PDF n = PDFs n = 0 The pdf of ˆp when n = 1 1

13 PDF n = PDFs n = 0 1/2 The pdf of ˆp when n = 2 1

14 PDF n = PDFs n = 0 1/3 2/3 The pdf of ˆp when n = 3 1

15 PDF n = PDFs n = 0 1/4 2/4 3/4 The pdf of ˆp when n = 4 1

16 PDF n = PDFs n = 0 1/5 2/5 3/5 4/5 The pdf of ˆp when n = 5 1

17 PDF n = PDFs n = 0 1/6 2/6 3/6 4/6 5/6 The pdf of ˆp when n = 6 1

18 PDF n = PDFs n = 0 1/8 2/8 3/8 4/8 5/8 6/8 7/8 The pdf of ˆp when n = 8 1

19 PDF n = PDFs n = 0 2/10 4/10 6/10 8/10 The pdf of ˆp when n = 10 1

20 PDF n = PDFs n = 0 2/12 4/12 6/12 8/12 10/12 The pdf of ˆp when n = 12 1

21 PDF n = PDFs n = 0 3/15 6/15 9/15 12/15 The pdf of ˆp when n = 15 1

22 PDF n = PDFs n = 0 4/20 8/20 12/20 16/20 The pdf of ˆp when n = 20 1

23 PDF n = PDFs n = 0 5/30 10/30 15/30 20/30 25/30 The pdf of ˆp when n = 30 1

24 PDF n = 180 PDFs n = The pdf of ˆp when n = 180

25 PDF n = 180 Normal curve PDFs n = The pdf of ˆp when n = 180

26 Outline PDFs n = 1 2 PDFs n =

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

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

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

30 Outline PDFs n = 1 2 PDFs n =

31 PDFs n = Theorem ( ) For any population, the sampling distribution of ˆp has the following mean and standard deviation: µˆp = p p(1 p) σˆp =. n Furthermore, the sampling distribution of ˆp is approximately normal, provided n is large enough.

32 PDFs n = The Sample Size n is large enough if np 5 and n(1 p) 5.

33 Outline PDFs n = 1 2 PDFs n =

34 PDFs n = Suppose that the true proportion of Americans who strongly approve of President Obama s permance is 30%. Can we use a sample of 1500 Americans to disprove the hypothesis that the rate is 35%?

35 Hypothesis Testing PDFs n = Let us test the hypotheses H 0 : 35% of all Americans strongly approve of President Obama s permance. H 1 : Less than 35% of all Americans strongly approve of President Obama s permance.

36 Hypothesis Testing PDFs n = Using the Central, the null hypothesis predicts that the distribution of ˆp is Normal, with p, which is 0.35 (hypothetically). (0.35)(0.65) Standard deviation 1500 = That is, ˆp is N(0.35, ). Design a decision rule so that α = What is the value of β?

37 Outline PDFs n = 1 2 PDFs n =

38 PDFs n = Read Sections , pages Exercises 7-14, page 526.

39 PDFs n = Answers 8. (a) (b) Yes. The probability of that is (a) The sampling distribution the sample proportion of men is normal with mean 0.49 and standard deviation (b)

40 PDFs n = Answers 12. (a) Draw a normal curve with mean 0.60 and standard deviation (b) (i) and (ii) and (iii) and (c) No. According to part (b)(ii), a proportion as low as 0.50 has only about a % chance of occurring. (d) Sketch a normal curve with mean 0.60 and standard deviation (i) It is narrower. (ii) They will each be half as wide as bee.

41 PDFs n = Answers 14. (a) It is normal with mean 0.50 and standard deviation (b) ˆp = p-value of 0.56 is (c) Reject H 0. A majority of shoppers favor longer shopping hours.