Lecture 31 Sections 9.4. Tue, Mar 17, 2009

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1 s for s for Lecture 31 Sections 9.4 Hampden-Sydney College Tue, Mar 17, 2009

2 Outline s for

3 s for Exercise 9.17, page 582. It is believed that 20% of all university faculty would be willing to work fewer hours for less pay in order to obtain more time for personal and leisure activities. Grace, the dean at a prominent university, thinks that among the female faculty this proportion would be 40%. Grace would like to test the following hypotheses about p, the population proportion of all female faculty who would be willing to work fewer hours for less pay in order to obtain more time for personal and leisure activities. H 0 : p = 0.20 versus H 1 : p = 0.40.

4 s for Exercise 9.17, page 582. Natalie, the director for the Institute of Social Research, took a random sample of female faculty from this university. Of the 600 respondents Natalie found that 45% were wiling to work fewer hours for less pay to obtain more personal and leisure time.

5 s for Exercise 9.17, page 582. (a) Give the estimate of the true proportion of female faculty who are willing to work fewer hours for less pay to obtain more time for personal and leisure activities. (b) Calculate the p-value of the test. (c) Give your decision using a 5% level of significance. (d) Calculate the power of the test.

6 s for Solution (a) estimate of the proportion of female faculty willing to work less is ˆp = (b) Use ˆp = 0.45 as the observed value. value of the test statistic is So the p-value is z = (0.20)(0.80) 600 = normalcdf(15.3,e99) = (c) Reject H 0.

7 s for Solution (d) This gets involved. power of the test is 1 β, so we must find β. β is the probability of a Type II error. That is, β is the probability that ˆp falls in the acceptance region when, in fact, H 1 is true. That means that we need to determine the acceptance region, i.e., we must find the critical value.

8 s for Solution (d) Since the level of significance is 5%, the critical value is the value that cuts off the upper 5% of the null distribution, which is (0.20)(0.80) N(0.20, ) = N(0.20, ). 600 So, the critical value is c = invnorm(0.95,0.20,0.163) =

9 s for Solution (d) Now we can say that β is P (ˆp < ), given that H 1 is true. alternative distribution is (0.40)(0.60) N(0.40, ) = N(0.40, 0.02). 600 So, β = normalcdf(-e99,0.2269,0.40,0.02) = 0. Finally, the power of the test is 1 β = 1 = 100%.

10 s for Portrait of Shakespeare unveiled. What does it mean to be 90% sure of something? 90% means 9 out of 10 of something. In this case, 9 out of 10 what?

11 s for Portrait of Shakespeare unveiled. What does it mean to be 90% sure of something? 90% means 9 out of 10 of something. In this case, 9 out of 10 what?

12 s for Portrait of Shakespeare unveiled. What does it mean to be 90% sure of something? 90% means 9 out of 10 of something. In this case, 9 out of 10 what?

13 s for Portrait of Shakespeare unveiled. What does it mean to be 90% sure of something? 90% means 9 out of 10 of something. In this case, 9 out of 10 what?

14 Point s for Definition (Point estimate) A point estimate is a single number that estimates the value of the parameter. x is a point estimator for µ. ˆp is a point estimator for p. s 2 is a point estimator for σ 2. s is a point estimator for σ.

15 Point s for problem with point estimates is that we have no idea how close we can expect them to be to the parameter. That is, they provide no idea of how large the error may be.

16 s for Definition ( estimate) An interval estimate is an interval of numbers that has a specified probability of containing the value of the parameter. Definition (Margin of error) margin of error of an interval estimate is half the width of the interval. An interval estimate is more informative than a point estimate.

17 s for x ± (margin of error) is an interval estimate for µ. ˆp ± (margin of error) is an interval estimate for p. s 2 ± (margin of error) is an interval estimate for σ 2. s ± (margin of error) is an interval estimate for σ. Question How do we find the margin of error? It depends on the level of confidence.

18 Levels s for Definition ( level) confidence level of an interval estimate is the probability that the interval contains the value of the parameter. If the confidence level is 95%, then the interval estimate is called a 95% confidence interval. Treating the center of the interval estimate as the point estimate, we are 95% sure that the value of the parameter is with the margin of error from the point estimate.

19 95% s s for Given the sample size n, the sampling distribution of ˆp is normal with mean p and standard deviation p(1 p). n refore, 95% of the time, ˆp will lie within 1.96 of p. p(1 p) n

20 s for Consider an analogy of a shooter shooting at a target. Suppose a shooter hits within 5 rings (5 inches) of the bull s eye 95% of the time. n each individual shot has a 95% chance of hitting within 5 inches.

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27 s for On any given shot, the shooter can be 95% sure that he hit the target.

28 s for Now suppose we are shown where the shot hit, but we are not shown where the bull s eye is. What is the probability that the bull s eye is within 5 inches of that shot?

29 s for Where is the bull's eye?

30 s for 95% Margin of error 5"

31 s for In a similar way, 95% of the sample proportions ˆp should lie within 1.96 standard deviations (σˆp ) of the parameter p.

32 s for p

33 s for 1.96σ p

34 s for 1.96σ p

35 s for 1.96σ p

36 s for 1.96σ p

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38 s for 1.96σ p

39 s for refore, if we compute a single ˆp, then there is a 95% chance that it lies within a distance 1.96σˆp of p.

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42 s for Where is p?

43 s for 95% Margin of error

44 Approximate 95% s s for Thus, the 95% confidence interval would be except... ˆp ± 1.96σˆp,

45 Approximate 95% s s for formula for σˆp is σˆp = p(1 p), n which requires that we know p, the very thing we are trying to estimate!

46 Approximate 95% s s for best we can do is to use ˆp in place of p to estimate σˆp : ˆp(1 ˆp) ˆp ± n

47 s for (95% confidence interval for p) Suppose we observe that of 1000 births, 520 of them are male. Estimate, using a 95% confidence interval, the true proportion of male births.

48 s for (95% confidence interval for p) point estimate is ˆp = = margin of error is ˆp(1 ˆp) (0.52)(0.48) 1.96 = 1.96 = n 1000 confidence interval is ˆp(1 ˆp) ˆp ± 1.96 = 0.52 ± n

49 s for (95% confidence interval for p) point estimate is ˆp = = margin of error is ˆp(1 ˆp) (0.52)(0.48) 1.96 = 1.96 = n 1000 confidence interval is ˆp(1 ˆp) ˆp ± 1.96 = 0.52 ± n

50 s for (95% confidence interval for p) point estimate is ˆp = = margin of error is ˆp(1 ˆp) (0.52)(0.48) 1.96 = 1.96 = n 1000 confidence interval is ˆp(1 ˆp) ˆp ± 1.96 = 0.52 ± n

51 s for (95% confidence interval for p) We are 95% confident that the point estimate of 0.52 is in error by no more than

52 s for Read Section 9.4, pages Let s Do It! 9.5. Exercises 19, 20, 21, 24, 29, page 595.

53 s for Answers 20. (a) 0.36 ± (b) 95% confidence means that if we repeated this procedure, including taking new samples, 95% of our confidence intervals would contain p and 5% would not. (c) It would not change. (d) margin of error would be smaller.

54 s for Answers 24. (a) (i) Yes. All telephone numbers were equally likely. It may not have been a simple random sample, because not all homes have telephones. (ii) population was intended to be all households in the Ann Arbor area, but they sampled only from those houses with telephones. (iii) Probably the poorer people are the ones without telephones and they are probably also the ones who use public transportation the most. refore, this could significantly bias the results. (b) ˆp = (c) 0.81 ± (d) margin of error is We could select a larger sample or we could lower the confidence level.

Lecture 21 Section Fri, Oct 3, 2008

Lecture 21 Section Fri, Oct 3, 2008 Lecture 21 Section 6.3.1 Hampden-Sydney College Fri, Oct 3, 2008 Outline 1 2 3 4 5 6 Exercise 6.15, page 378. A young woman needs a 15-ampere fuse for the electrical system in her apartment and has decided