Confidence Intervals and Hypothesis Testing

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Motivational Scenario. A market research agency has been given the task to estimate the average number of hours per week that young adults spend surfing the web.The agency surveys a random sample of 100 young adults and obtains a mean of 20 hours and a standard deviation of 5 hoursCan the agency c

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Confidence Intervals and Hypothesis Testing

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1. Confidence Intervals and Hypothesis Testing 7.1/2 Confidence Intervals for a Population Mean 7.4 Confidence Intervals for a Population Proportion 8.1 Null and Alternative Hypotheses and Errors in Testing 8.2/3 Tests about a Population Mean assuming sigma is known; use of Z 8.4 Testing assuming sigma is unknown

2. Motivational Scenario A market research agency has been given the task to estimate the average number of hours per week that young adults spend surfing the web. The agency surveys a random sample of 100 young adults and obtains a mean of 20 hours and a standard deviation of 5 hours Can the agency conclude that the true mean number of hours per week spent by all young adults surfing the web is exactly 20 hours?

3. Motivational Scenario – cont’d Because the market research agency recognizes that the 20 hours was obtained from just one of many possible samples of the population they are unwilling to say the population mean is exactly equal to 20 hours. To allow for the variation in the sample estimate they may cautiously conjecture that the true mean is somewhere between 18 and 22 hours, between 15 and 25 hours, etc.

4. Establishing an Interval for Estimation How wide should they make the interval? How confident should they be that the named interval does indeed contain the true mean? On what basis should the choice be made? They can use an established fact about how sample means vary when random samples are repeatedly drawn from any population – the central limit theorem

5. Confidence Intervals for the Mean - Rationale

8. 7.1 Confidence Intervals for a Mean When Sigma is Known

9. Example: Confidence Interval for Mean assuming sigma known

10. Effect of Confidence Coefficient on Interval

11. Effect of Sample Size on Confidence Interval

13. The t-distribution Just as has a standard normal (Z) distribution either when X is normal or n is large, so does follow a t-distribution with n-1 degrees of freedom provided X is normally distributed

14. The t-distribution

16. The t-distribution

17. Effect of Degrees of Freedom on the t-distribution

18. 7.2 Confidence Intervals for a Mean when Sigma is unknown

20. 7.4 Confidence Intervals for a Population Proportion

21. Example: Confidence Interval for a Proportion

22. Hypothesis Testing Formal way to determine whether or not the data support a belief or hypothesis.

23. Hypothesis Testing in the Judicial System In our judicial system we have the following hypotheses: The accused is innocent; The accused is guilty We can make two errors: Convicting the innocent; Letting the guilty go free

24. Hypothesis Testing in the Judicial System It is desirable to minimize the chance of committing either error. But guarding against one usually results in increasing the chance of committing the other. Society favors guarding against convicting the innocent.

25. Procedure for Guarding Against Convicting Innocent Assume accused is innocent. Gather evidence to prove guilt Convict only if evidence is strong enough

26. Equivalent Procedure in Statistics To test the belief m > 20 (alternative hypothesis, Ha) Assume m not > 20; (Null hypothesis, Ho: m ? 20) Gather random sample from population & compute sample mean, x Conclude m > 20 (Ha) only if evidence is strong enough, i.e. if x is so many standard deviations away from 20, the probability of this occurring by pure random chance is very small

27. 8.1 Null and Alternative Hypotheses

28. Types of Hypothesis

31. Formal Hypothesis Testing Establish the Null Hypothesis and the Alternative Hypothesis : H0: ? = $20.00 Ha: ? ? $20.00 (two tailed) OR H0: ? = $20.00 Ha: ? > $20.00 (right tailed) OR H0: ? = $20.00 Ha: ? < $20.00 (left tailed) Ho must always have an equal sign and Ha must be what you want to prove

32. Formal Hypothesis Testing for m Specify ?, the probability of a Type I error. For example, ?=.05. You set your standard for how extreme the sample results must be (in support of the alternative hypothesis in order for you to reject the null. Here, the sample results must be strong enough in favor of Ha that you would falsely reject the null only 5% of the time.

33. Formal Hypothesis Testing for m Compute the test statistic, z. a) Determine the rejection point for z, ±za b) Compute the probability of obtaining such an extreme test statistic z by pure chance, if the Null hypothesis were true. This is called the p-value of the test. Reject or fail to reject the Null Hypothesis by determining a) whether |z| > za or b) the p-value is less than or greater than a. Interpret statistical result in (real-world) managerial terms

34. Interpreting p-values All statistical packages give p-values in the standard output. When we reject Ho we say the test is significant. If p-value < .01, highly significant (overwhelming evidence in support of research hypothesis) If p-value between .01 and .05, significant (strong evidence) If p-value between .05 and .10, slightly significant (weak evidence) If p-value > .10, not significant (no evidence)

35. Hypothesis Test Example; ex. 8.7/8.30 What are we given? n = 65; s = 2.6424; xbar = 42.954; For this exercise we’ll consider s to represent the true s Step 1, establish hypotheses H0: ? = 42 vs. Ha: ? > 42 Step 2, specify significance level. a = .01 (given) Step 3, compute the test statistic z = (42.954-42)/(2.6424/sqrt65) = 2.91 Step 4a, determine the rejection point, z..01 = 2.326 Step 4b, determine the p-value. Z-table gives P(z < 2.91) = 0.9982. So, P(z > 2.91) = 1- 0.9982 = .0018

36. Hypothesis Test Example; ex. 8.7/8.30 Step 5, decision; reject Ho since a) test statistic, z (2.91) > rejection point (2.326) or since b) p-value (.0018) < ? = .01 Step 6, conclusion within context: Conclude there is very strong evidence that a typical customer is very satisfied since we reject the notion that the average rating is = 42 with such a small p-value

37. Hypothesis Test Example; ex. 8.9/8.32 What are we given? n = 100; s = 32.83; xbar = 86.6; For this exercise we’ll consider s to represent the true s Step 1, establish hypotheses H0: ? = 90 vs. Ha: ? < 90 Step 2, specify significance level. a = .05 (arbitrary) Step 3, compute the test statistic z = (86.6 - 90)/(32.83/sqrt100) = -1.035 Step 4a, determine the rejection point, -z.05 = -1.645 Step 4b, determine the p-value. Z-table gives P(z < -1.04) = 0.1492.

38. Hypothesis Test Example; ex. 8.9/8.32 Step 5, decision; fail to reject Ho since a) test statistic, z (-1.035) not beyond rejection point (-1.645) or since b) p-value (.1492) > ? = .05 Step 6, conclusion within context: Conclude there is no evidence that the mean audit delay is less than 90 days since we fail to reject the notion that the mean is = 90

39. Two Tailed vs. One Tailed Tests

40. Hypothesis Test Example; ex. 8.11/8.45a What are we given? n = 36; s = 0. 1; xbar = 16.05; For this exercise we’ll consider s to represent the true s Step 1, establish hypotheses H0: ? = 16 vs. Ha: ? ? 16 Step 2, specify significance level. a = .01 (given) Step 3, compute the test statistic z = (16.05 - 16)/(0.1/sqrt36) = 3.0 Step 4a, determine the rejection points, ±z.005 = ±2.576 Step 4b, determine the p-value. Z-table gives P(z < 3) = 0.9987, P(z > 3) = 1 - .9987 = 0.0013. So 2-tailed p-value = 2*0.0013 = 0.0026

41. Hypothesis Test Example; ex. 8.11/8.45 Step 5, decision; reject Ho since a) test statistic, z 3.0) is beyond one of the rejection points (+2.576) or since b) p-value (.0026) < ? = .01 Step 6, conclusion within context: Conclude there is very strong evidence that the filler needs readjusting since we reject the notion that the mean is = 16

42. 8.4 Tests about a Population Mean; Sigma Unknown

43. Example 1: Test about a Mean; Sigma Unknown. Credit Card Case (pg 322) What are we given? n = 15; s = 1.538; xbar = 16.827; ? = .05; Step 1, establish hypotheses H0: ? ? 18.8 vs Ha: ? < 18.8 Step 2, set significance level. a = .05 (given) Step 3, compute the test statistic Step 4a, determine rejection point; –t.05,14 = –1.761 Step 4b, estimate p-value; df=14, P(t < –4.14) = .0005. So P(t < –4.97) < .0005

44. Example 1: Test about a Mean; Sigma Unknown. Credit Card Case (pg 322)

45. MegaStat Output for Example 1

46. Hypothesis Test Example 2 What are we given? n = 30; s = 15; x = 21; ? = .10; Step 1, establish hypotheses H0: ? = 20 vs. Ha: ? > 20 Step 2, set significance level. a = .10 (given) Step 3, compute the test statistic t = (21 - 20)/2.74 = 0.365 Step 4a, determine the rejection points, t.10,29 = 1.311 Step 4b, estimate the p-value. Using df = 29, t-table gives P(T > 1.311) = .10. So, P(T > t = .365) > .10 Best estimate of p-value is > 0.10

48. Hypothesis Test Example 2 Step 5, decision; fail to reject Ho since (a) test statistic, t (0.365) < rejection point (1.311) or (b) p-value (>.10) > ? = .10 Step 6, conclusion within context: no context given but we can say there is insufficient evidence that the population mean is greater than 20. Notice we do NOT say we have evidence that m is less than or equal 20. In other words we can prove Ha but not Ho

49. MegaStat Output for Example 2

50. Hypothesis Test Example 3 What are we given? n = 400; s = 15; x = 23; ? = .05; Step 1, establish hypotheses H0: ? = 20 vs. Ha: ? ? 20 Step 2, set significance level. a = .05 (given) Step 3, compute the test statistic t = (23 – 20)/0.75 = 4.0 Step 4a, determine the rejection points, ±t.025,399 = ±1.96 Step 4b, estimate the p-value. Using df = 8, t-table gives P(T > 3.291) = .0005. So, P(T > t = 4) < .0005. Since test is two tailed, p-value < 2*0.0005 i.e. < 0.001

52. Hypothesis Test Example 3 Step 5, decision; reject Ho since (a) test statistic, t (4.0) > rejection point (1.96) or (b) p-value (<.001) < ? = .05 Step 6, conclusion within context: no context given but we can say there is very strong evidence that the population mean is not equal to 20 and is most likely > 20 since we rejected Ho at the upper tail.

53. MegaStat Output for Example 3

54. Hypothesis Test Example 4 What are we given? n = 25; s = 4; x = 18.7; ? = .05; Step 1, establish hypotheses H0: ? = 20 vs. Ha: ? < 20 Step 2, set significance level. a = .05 (given) Step 3, compute the test statistic t = (18.7 – 20)/0.80 = – 1.63 Step 4a, determine the rejection point, – t.05,24 = –1.711 Step 4b, estimate the p-value. Using df = 24, t-table gives P(T < –1.711) = .05 and P(T < –1.318) =.10 Since – 1.711 < (t = –1.63) < –1.318 , p-value is between 0.05 and 0.10

56. Hypothesis Test Example 4 Step 5, decision; F.T.R. Ho since (a) test statistic, t (–1.63) is not as extreme as rejection point (–1.711) or (b) p-value (between .05 and .10) > ? = .05 Step 6, conclusion within context: no context given but we can say there is only weak evidence that the population mean is less than 20 and the finding is insignificant at the 5% level

57. MegaStat Output for Example 4

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