1. Chapter 6.1 Estimating with Confidence 1

2. Point estimation  Sample mean is the natural estimator of the unknown population mean.  Is the point estimation a good method? 1. It may never hit the true value (population mean). 2. We have no idea about the variability of the estimation. Therefore, we have no confidence about how close our estimator is to the true value. Taser GunNet Gun Net Gun (flyswatter)  Idea: It is better to use an INTERVAL than a POINT estimator.

3. Review: Chapter 1.3: All Normal curves N  ) share 68-95-99.7 Rule Reminder: µ (mu) is the mean of the idealized curve, while x¯ is the mean of a sample. s (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample. About 68% of all observations are within 1 SD (  of mean (  ). Called: C=68%, z*≈1 About 95% of all observations are within 2  of the mean . Called: C=95%, z* ≈ 2 Almost all (99.7%) observations are within 3  of the mean. Called: C=99.7%, z* ≈ 3 Standard Normal Distribution N(0, 1)

4. Confidence intervals contain the population mean  in C% of samples. Different areas under the curve give different confidence levels C. Example: For an 80% confidence level C, 80% of the normal curve’s area is contained in the interval. C z*z*−z*−z* Confidence levels z*:  z* is related to the chosen confidence level C.  C is the area under the standard normal curve between −z* and z*. The confidence interval is thus:

5. Point estimation versus interval  When population mean (µ) is unknown, it is better to use an interval than a point to estimate it.  The theory behind interval estimation looks at the sampling distribution of the statistic.  Confidence level C- CI for the population mean µ is : For a particular confidence level, C, the appropriate z* value is given in the last row of Table D. Example: For a 98% confidence level, z*=2.326

6. Specific Confidence Intervals for population mean 99% CI for the population mean µ is : i.e.: C=99%, z*=2.576 95% CI for the population mean µ is : i.e.: C=95%, z*=1.960 90% CI for the population mean µ is : i.e.: C=90%, z*=1.645

7. Example 1  The average lifetime of 36 randomly selected certain brand TVs is 20 years. Suppose the SD of all TVs is 2 years.  Construct a 95% CI for the average lifetime of all TVs from this brand. A 95% CI for the average lifetime of all TVs from this brand is: (19.35, 20.65)

8. Example 2 1. The average height of 100 randomly selected UNCW students is 5.9 feet. Suppose the SD of the heights of all students is 1.2 feet.  Construct 99%, 95% and 90% CIs for the average height of all students. 2. Select another set of 100 UNCW students randomly. The average height of second set of 100 students is 5.5 feet. Suppose the SD of the heights of all students is 1.2 feet.  Construct 95% CIs for the average height of all students. A 99% CI for the average height of all students is: (5.5904, 6.2096) A 95% CI for the average height of all students is: (5.6648, 6.1352) A 90% CI for the average height of all students is: (5.702, 6.098) A 95% CI for the average height of all students is: (5.2648, 5.7352) Note: Confidence level C gets smaller, CI gets smaller

9. Outlines for Z*  Z* depends on the level of confidence C.  What does “confidence” mean? See applet.applet.  This idea is only true for simple random samples and completely randomized experiments.  Margin of error: Z*  /√(n)  http://bcs.whfreeman.com/ips7e/#616906__657132__ http://bcs.whfreeman.com/ips7e/#616906__657132__  In Statcrunch, use Applet option and select desired options. Then run the applet.  http://www.statcrunch.com/app/index.php? http://www.statcrunch.com/app/index.php

10. Understanding of Confidence Intervals With 95% confidence, we can say that µ should be within roughly 2 standard deviations (that is, 2*  /√n) from our sample mean.  About 95% of all possible samples of this size n, µ will indeed fall in our confidence interval.  About only 5% of samples would be farther from µ.  applet. applet.

11. Link between confidence level and margin of error The margin of error depends on z. C z*z*−z* m Higher confidence C implies a larger margin of error m (thus less precision in our estimates). A lower confidence level C produces a smaller margin of error m (thus better precision in our estimates).

12. Example 3  a)A 90% CI for the average life that this medicine can prolong for all cancer patients is: ( 3.8453, 4.1547); b)Z*=1.645; MOE=(1.645)*(0.75)/sqrt(n)=0.1; so n=(1.645*0.75/0.1)^2=152.2139. We will take n=153.

13. Summary to Confidence Interval  If Confidence level C gets larger and n stays the same, what will happen to z*, MOE, CI, and prediction precision?  If Z* and  stay the same, when n goes bigger, what will happen to MOE and CI?

14. Chapter 6.2 Hypothesis Testing

15. 6.2 Tests of hypothesis 5 Steps to Hypothesis Testing 1. State the hypothesis 2. State the level of significance 3. Calculate the test statistic 4. Find the p-value 5. Conclusion (both statistical and non-statistical)

16. Hypothesis Testing  The idea of hypothesis testing is to use the data to make a decision. In hypothesis testing, there are only two “decisions”, also called hypotheses, in which the data could support. The two hypotheses are called the null hypothesis and the alternative hypothesis.  Forms of Null and Alternative Hypotheses

17. Null Hypothesis Expectation --- what somebody believes or claims before the sample available.  Null hypothesis: the hypothesis you assume to be true, the one you are comparing against your data. denoted by H 0.  Many times the null hypothesis is a statement of “no effect” or of “no difference”…  “Being fair”  E.g.: Last year, your company’s service technicians took an average of 2.6 hours to response to trouble calls from business customers who had purchased service contracts. Do this year’s data show a lower average response time?

18. Alternative Hypothesis Expectation is not correct --- the difference between the expectation and sample statistic is real.  Alternative hypothesis: express the hopes or suspicions we bring to data.  The test is designed to assess the strength of evidence against the null hypothesis, denoted by H a.  It’s a statement that “supports” the information from the data.  E.g.: Last year, your company’s service technicians took an average of 2.6 hours to response to trouble calls from business customers who had purchased service contracts. Do this year’s data show a lower average response time?

19. Example 1  Exercise 6.55 (p. 391): Translate each of the following research questions into appropriate H 0 and H a. a) Census Bureau data show that the mean household income in the area served by a shopping mall is $62,500 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population. a) H 0 : µ = $62,500 verse H a : µ > $62,500.

20. Example 1 cont.  Exercise 6.55 (p. 391): Translate each of the following research questions into appropriate H 0 and H a. b) Last year, your company’s service technicians took an average of 2.6 hours to response to trouble calls from business customers who had purchased service contracts. Do this year’s data show a different average response time? b) H 0 : µ = 2.6 hours verse H a : µ ≠ 2.6 hours.

21. 6.2 Tests of hypothesis 5 Steps to Hypothesis Testing 1. State the hypothesis 2. State the level of significance (α=0.05 unless otherwise stated) 3. Calculate the test statistic 4. Find the p-value 5. Conclusion (both statistical and non-statistical)

22. 6.2 Tests of hypothesis--Review E.g.: Last year, your company’s service technicians took an average of 2.6 hours to response to trouble calls from business customers who had purchased service contracts. Do this year’s data show a lower average response time? E.g.: Census Bureau data show that the mean household income in the area served by a shopping mall is $62,500 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population. E.g.: Last year, your company’s service technicians took an average of 2.6 hours to response to trouble calls from business customers who had purchased service contracts. Do this year’s data show a different average response time? b) H 0 : µ = $62,500 verse H a : µ > $62,500. c) H 0 : µ = 2.6 hours verse H a : µ ≠ 2.6 hours. a) H 0 : µ = 2.6 hours verse H a : µ < 2.6 hours.

23. One-sided and two-sided tests for P-value  A two-tail or two-sided test of the population mean has these null and alternative hypotheses: H 0 : µ = [a specific number] H a : µ  [a specific number]  A one-tail or one-sided test of a population mean has these null and alternative hypotheses: H 0 : µ = [a specific number] H a : µ < [a specific number] OR H 0 : µ = [a specific number] H a : µ > [a specific number]

24. Find P-value The P-value is the area under the sampling distribution for values at least as extreme, in the direction of H a, as that of our random sample. Use Table A, or NORMALCDF in calculator. e.g. H 0 : µ = 2.6 hours verse H a : µ < 2.6 hours gives test statistic Z=-1.6. Q: Find the p-value. µ defined by H 0 Sampling distribution σ/√n

25. P-value in one-sided and two-sided tests To calculate the P-value for a two-sided test, use the symmetry of the normal curve. Find the P-value for a one-sided test, and double it. One-sided (one-tailed) test Two-sided (two-tailed) test

26. Example 3: One-sample Z-test A test of the null hypothesis H 0 : µ = µ 0 gives test statistic Z=-1.6 a) What is the P-value if the alternative is H a : µ > µ 0 ? b) What is the P-value if the alternative is H a : µ < µ 0 ? c) What is the P-value if the alternative is H a : µ ≠ µ 0 ?

27. Example 3(cont.): One-sample Z-test A test of the null hypothesis H 0 : µ = µ 0 gives test statistic Z=2.1 a) What is the P-value if the alternative is H a : µ > µ 0 ? b) What is the P-value if the alternative is H a : µ < µ 0 ? c) What is the P-value if the alternative is H a : µ ≠ µ 0 ?

28. How to do 5 steps 1. State H 0 and H a 2. State the level of significance (Usually α is 5% ). 3. Calculate the test statistic (ASSUMING THE NULL HYPOTHESIS IS TRUE) 4. Find the P-value, that is the probability in the direction of H a. 5. Draw Conclusion: If P-value ≤ α, then we reject H 0 (Enough evidence). If P-value > α, then we do not reject H 0 (No Enough evidence). Note: The two possible conclusions are rejecting or not rejecting H 0.

29. Example 2: The P-value for a significance test is 0.032 a) Do you reject the null hypothesis at level α = 0.05? b) Do you reject the null hypothesis at level α = 0.01? c) Explain your answers. Note that: If P-value ≤ α, then we reject H 0 (Enough evidence). If P-value > α, then we do not reject H 0 (No Enough evidence).

30. 30 Chap 5: Sampling distribution of a sample mean=distribution of Population

31. One-sample Z-test for population mean:  Test statistics is a Z-score to the sampling distribution of the sample mean (see chapter 5.1)

32. Example 4: One-sample Z-Test (one sided)  The National Center for Health Statistics reports that the mean systolic blood pressure for males 35 to 44 years of age is 128 with a population SD=15.  The medical director of a company looks at the medical records of 72 company executives in this age group and finds that the mean systolic blood pressure in this sample is 126.07. Is this evidence that executives blood pressures are lower than the national average?

33. Answer to Example 4: (1) Hypothesis: H 0 : µ = 128 v.s. H a : µ <128. (2) α = 5% (3) One-sample Z-Test statistics (4) From table A, the area under the standard normal curve to the left of z is 0.1379.Thus, P-value = normalcdf(-999, -1.09, 0, 1) = 0.1379 = 13.79%. (5) (Statistical Conclusion) Since P-value > α, we do not reject H0. (Non- Statistical Conclusion) That is, there is NO evidence that executives blood pressures are lower than the national average.

34. Example 5: One-sample Z-Test (two sided) A new medicine treating cancer was introduced to the market decades ago and the company claimed that on average it will prolong a patient’s life for 5.2 years. Suppose the SD of all cancer patients is 2.52. In a 10 years study with 64 patients, the average prolonged lifetime is 4.6 years. With normality assumption, do the 10-year study’s data show a different average prolonged lifetime?

35. Answer to Example 5: (1) Hypothesis: H 0 : µ = 5.2 year versus H a : µ ≠ 5.2 year. (2) α = 5% (3) One-sample Z-Test statistics (4) From table A, the area under the standard normal curve to the left of z=- 1.90 is normalcdf(-999, -1.90, 0, 1) = 0.0287.Thus, P-value = 2*0.0287 = 0.0574=5.74%. (Statistical Conclusion) Since P-value > α, we do not reject H0. (Non-Statistical Conclusion) There is NOT enough evidence to conclude that 10-year study’s data show a different average prolonged lifetime.

36. Example 5.1 (Based on Example 5) Find a 95% confidence interval for the average prolonged lifetime for all patients. A 95% CI for the average prolonged lifetime for all patients is given by: [3.9826, 5.2174] Note: Since H 0 : µ =5.2, we have µ 0 =5.2 which falls inside the 95% CI. We are therefore 95% confident that µ is equal to 5.2. Therefore, we did not reject H 0 at the level of 5%.

37. Confidence intervals to test hypotheses For a level  two-sided significance test: Rejects H 0 :  =  0 exactly when the hypothesized value  0 falls outside a level (1-  confidence interval for . α / 2 In a two-sided test, C = 1 – α. C confidence level α significance level

38. Ex: Your sample gives a 99% confidence interval of. With 99% confidence, could samples be from populations with µ = 0.86? µ = 0.85? 99% C.I. Logic of confidence interval test Cannot reject H 0 :  = 0.85 Reject H 0 :  = 0.86

39. Example 6: The P-value for a two-sided test of the null hypothesis H 0 : µ = 30 is 0.04. a) Does the 95% confidence interval include the value 30? Why? b) Does the 90% confidence interval include the value 30? Why? c) Does the 99% confidence interval include the value 30? Why?  Note that In a two-sided test, C = 1 – α. C confidence level, α significance level

40. Example 7: A 90% confidence interval for a population mean is (12, 15). a) Can you reject the null hypothesis that H 0 : µ = 13 at the 10% significance level? Why? b) Can you reject the null hypothesis that H 0 : µ = 10 at the 10% significance level? Why?

41. Multiple Choice Questions 10. The P-value for a two-sided test of the null hypothesis is 0.09, a) the 99% confidence interval includes the value 30. b) the 95% confidence interval includes the value 30. c) the 90% confidence interval does not include the value 30. d) All of the above are correct.

42. Exercises on Hypothesis Testing (One-sample Z-test) 1. Because of variation in the manufacturing process, tennis balls produced by a particular machine do not have identical diameters, which is supposed to be 3in. The population SD is 0.15 in. If the average diameters of the first 36 balls made from a machine is 3.2in, shall we stop and calibrate the machine? 2. A new medicine treating cancer was introduced to the market decades ago and the company claimed that on average it will prolong a patient’s life for 5 years. The population SD is 0.4 year. In a 10 years study with 81 patients, the average prolonged lifetime is 4.5 years. With normality assumption, shall we reject the original claim? 3. The registrar office claims that the average SAT score of UNCW students is 1050. The population SD is 80. Suppose you randomly select 100 UNCW students the SAT score average of your sample is 1020. Do you agree with the claim? 4. National data shows that on the average, college freshmen spend 7.5 hours a week going to parties. The population SD is 2 hours. One administrator takes a random sample of 81 freshmen from her college and finds out that her students’ average hours spent on parties is 6.6. Shall the administrator believe that the national data applies to her students?

43. Solutions 1. H 0 : µ = 3, H a : µ ≠ 3; α = 5%;Z=(3.2-3)/(.15/(36)^.5)=8; P-value = 2*0= 0<5%;we reject H0 and we shall stop and calibrate the machine. 2. H 0 : µ = 5, H a : µ ≠ 5; α = 5%;Z=(4.5-5)/(.4/(81)^.5)=-11.25; P-value = 2*0= 0<5%;we reject H0 and we shall reject the claim that the average is 5 years. 3. H 0 : µ = 1050,H a :µ ≠ 1050; α = 5%;Z=(1020-1050)/(80/(100)^.5)=-3.75; P-value = 2*0= 0<5%;we reject H0 and we shall stop and calibrate the machine. 4. H 0 : µ = 7.5, H a : µ ≠ 7.5; α = 5%;Z=(6.6-7.5)/(2 /(81)^.5)=-4.05; P-value = 2*0= 0<5%;we reject H0 and the national data does not apply.