1. Fundamentals of Hypothesis Testing: One-Sample Tests İŞL 276

2. What is a Hypothesis?  A hypothesis is a claim (assumption) about a population parameter: population mean population proportion Example: The mean monthly cell phone bill of this city is μ = $42 Example: The proportion of adults in this city with cell phones is π = 0.68

3. The Null Hypothesis, H 0  States the claim or assertion to be tested Example: Average number of mobile phone in a Turkish family is equal to 3 ( )  Is always about a population parameter, not about a sample statistic

4. The Null Hypothesis, H 0  Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until proven guilty  Refers to the status quo  Always contains “=”, “≤” or “  ” sign  May or may not be rejected (continued)

5. The Alternative Hypothesis, H 1  Is the opposite of the null hypothesis e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H 1 : μ ≠ 3 )  Challenges the status quo  Never contains the “=”, “≤” or “  ” sign  May or may not be proven  Is generally the hypothesis that the researcher is trying to prove

6. Population Claim: the population mean age is 50. (Null Hypothesis: REJECT Suppose the sample mean age is 20: X = 20 Sample Null Hypothesis 20 likely if μ = 50?  Is Hypothesis Testing Process If not likely, Now select a random sample H 0 : μ = 50 ) X

7. Sampling Distribution of X μ = 50 If H 0 is true If it is unlikely that we would get a sample mean of this value...... then we reject the null hypothesis that μ = 50. Reason for Rejecting H 0 20... if in fact this were the population mean… X

8. Level of Significance,   Defines the unlikely values of the sample statistic if the null hypothesis is true Defines rejection region of the sampling distribution  Is designated by , (level of significance) Typical values are 0.01, 0.05, or 0.10  Is selected by the researcher at the beginning  Provides the critical value(s) of the test

9. Level of Significance and the Rejection Region H 0 : μ ≥ 3 H 1 : μ < 3 0 H 0 : μ ≤ 3 H 1 : μ > 3   Represents critical value Lower-tail test Level of significance =  0 Upper-tail test Two-tail test Rejection region is shaded /2 0   H 0 : μ = 3 H 1 : μ ≠ 3

10. Hypothesis Tests for the Mean  Known  Unknown Hypothesis Tests for  (Z test) (t test)

11. Z Test of Hypothesis for the Mean (σ Known)  Convert sample statistic ( ) to a Z test statistic X The test statistic is: σ Knownσ Unknown Hypothesis Tests for   Known  Unknown (Z test) (t test)

12. Critical Value Approach to Testing  For a two-tail test for the mean, σ known:  Convert sample statistic ( ) to test statistic (Z statistic )  Determine the critical Z values for a specified level of significance  from a table or computer  Decision Rule: If the test statistic falls in the rejection region, reject H 0 ; otherwise do not reject H 0

13. Do not reject H 0 Reject H 0  There are two cutoff values (critical values), defining the regions of rejection Two-Tail Tests  /2 -Z 0 H 0 : μ = 3 H 1 : μ  3 +Z  /2 Lower critical value Upper critical value 3 Z X

14. 6 Steps in Hypothesis Testing 1. State the null hypothesis, H 0 and the alternative hypothesis, H 1 2. Choose the level of significance, , and the sample size, n 3. Determine the appropriate test statistic and sampling distribution 4. Determine the critical values that divide the rejection and nonrejection regions

15. 6 Steps in Hypothesis Testing 5. Collect data and compute the value of the test statistic 6. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the nonrejection region, do not reject the null hypothesis H 0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem (continued)

16. Hypothesis Testing Example Test the claim that the true mean of mobile phone in a Turkish family is equal to 3. Assume σ = 0.8, α=0.05 and sample results are n = 100, = 2.84 1. State the appropriate null and alternative hypotheses  H 0 : μ = 3 H 1 : μ ≠ 3 (This is a two-tail test) 2. Specify the desired level of significance and the sample size  Suppose that  = 0.05 and n = 100 are chosen for this test

17. Hypothesis Testing Example 3. Determine the appropriate technique  σ is known so this is a Z test. 4. Determine the critical values  For  = 0.05 the critical Z values are ±1.96 5. Collect the data and compute the test statistic  Suppose the sample results are n = 100,X = 2.84 (σ = 0.8 is assumed known) So the test statistic is: (continued)

18. Reject H 0 Do not reject H 0  6. Is the test statistic in the rejection region?  = 0.05/2 -Z= -1.96 0 Reject H 0 if Z 1.96; otherwise do not reject H 0 Hypothesis Testing Example (continued)  = 0.05/2 Reject H 0 +Z= +1.96 Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

19. 6(continued). Reach a decision and interpret the result -2.0 Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of phone in Turkish family is not equal to 3 Hypothesis Testing Example (continued) Reject H 0 Do not reject H 0  = 0.05/2 -Z= -1.96 0  = 0.05/2 Reject H 0 +Z= +1.96

20. p-Value Approach to Testing  p-value: Probability of obtaining a test statistic more extreme ( ≤ or  ) than the observed sample value given H 0 is true Also called observed level of significance Smallest value of  for which H 0 can be rejected

21. p-Value Approach to Testing  Convert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic )  Obtain the p-value from a table or computer  Compare the p-value with  If p-value < , reject H 0 If p-value  , do not reject H 0 X (continued)

22. 0.0228  /2 = 0.025 p-Value Example  Example: How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is  = 3.0? -1.960 -2.0 Z1.96 2.0 X = 2.84 is translated to a Z score of Z = -2.0 p-value = 0.0228 + 0.0228 = 0.0456 0.0228  /2 = 0.025

23.  Compare the p-value with  If p-value < , reject H 0 If p-value  , do not reject H 0 Here: p-value = 0.0456  = 0.05 Since 0.0456 < 0.05, we reject the null hypothesis (continued) p-Value Example 0.0228  /2 = 0.025 -1.960 -2.0 Z1.96 2.0 0.0228  /2 = 0.025

24. Connection to Confidence Intervals For sample mean is 2.84 and σ = 0.8 and n = 100, the 95% confidence interval is: 2.6832 ≤ μ ≤ 2.9968  Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at  = 0.05

25. One-Tail Tests  In many cases, the alternative hypothesis focuses on a particular direction H 0 : μ ≥ 3 H 1 : μ < 3 H 0 : μ ≤ 3 H 1 : μ > 3 This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3 This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3

26. Reject H 0 Do not reject H 0  There is only one critical value, since the rejection area is in only one tail Lower-Tail Tests  -Z 0 μ H 0 : μ ≥ 3 H 1 : μ < 3 Z X Critical value

27. Reject H 0 Do not reject H 0 Upper-Tail Tests  ZαZα 0 μ H 0 : μ ≤ 3 H 1 : μ > 3  There is only one critical value, since the rejection area is in only one tail Critical value Z X _

28. Ex:Upper-Tail Z Test for Mean (  Known) A phone industry manager thinks that customer monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test this claim. (Assume  = 10 is known and α=0.10) H 0 : μ ≤ 52 the average is not over $52 per month H 1 : μ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim) Form hypothesis test:

29. Reject H 0 Do not reject H 0  Suppose that  = 0.10 is chosen for this test Find the rejection region:  = 0.10 1.28 0 Reject H 0 Reject H 0 if Z > 1.28 Example: Find Rejection Region (continued)

30. Review:One-Tail Critical Value Z.07.09 1.1.8790.8810.8830 1.2.8980.9015 1.3.9147.9162.9177 z 01.28. 08 Standardized Normal Distribution Table (Portion) What is Z given  = 0.10?  = 0.10 Critical Value = 1.28 0.90.8997 0.10 0.90

31. Obtain sample and compute the test statistic Suppose a sample is taken with the following results: n = 64, X = 53.1 (  =10 was assumed known) Then the test statistic is: Example: Test Statistic (continued)

32. Reject H 0 Do not reject H 0 Example: Decision  = 0.10 1.28 0 Reject H 0 Do not reject H 0 since Z = 0.88 ≤ 1.28 i.e.: there is not sufficient evidence that the mean bill is over $52 Z = 0.88 Reach a decision and interpret the result: (continued)

33. Reject H 0  = 0.10 Do not reject H 0 1.28 0 Reject H 0 Z = 0.88 Calculate the p-value and compare to  (assuming that μ = 52.0) (continued) p-value = 0.1894 p -Value Solution Do not reject H 0 since p-value = 0.1894 >  = 0.10

34. t Test of Hypothesis for Mean (σ Unknown)  Convert sample statistic ( ) to a t test statistic X The test statistic is: Hypothesis Tests for  σ Knownσ Unknown  Known  Unknown (Z test) (t test)

35. Example: Two-Tail Test(  Unknown) The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $172.50 and s = $15.40. Test at the  = 0.05 level. (Assume the population distribution is normal) H 0 : μ  = 168 H 1 : μ  168

36.   = 0.05  n = 25   is unknown, so use a t statistic  Critical Value: t 24 = ± 2.0639 Example Solution: Two-Tail Test Do not reject H 0 : not sufficient evidence that true mean cost is different than $168 Reject H 0  /2=.025 -t n-1,α/2 Do not reject H 0 0  /2=.025 -2.0639 2.0639 1.46 H 0 : μ  = 168 H 1 : μ  168 t n-1,α/2

37. Hypothesis Tests for Proportions  Involves categorical variables  Two possible outcomes “Success” (possesses a certain characteristic) “Failure” (does not possesses that characteristic)  Fraction or proportion of the population in the “success” category is denoted by π

38. Proportions  Sample proportion in the success category is denoted by p  When both nπ and n(1-π) are at least 5, p can be approximated by a normal distribution with mean and standard deviation (continued)

39.  The sampling distribution of p is approximately normal, so the test statistic is a Z value: Hypothesis Tests for Proportions nπ  5 and n(1-π)  5 Hypothesis Tests for p nπ < 5 or n(1-π) < 5 Not discussed in this chapter

40.  An equivalent form to the last slide, but in terms of the number of successes, X: Z Test for Proportion X  5 and n-X  5 Hypothesis Tests for X X < 5 or n-X < 5 Not discussed in this chapter

41. Example: Z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the  = 0.05 significance level. Check: n π = (500)(.08) = 40 n(1-π) = (500)(.92) = 460

42. Z Test for Proportion: Solution  = 0.05 n = 500, p = 0.05 Reject H 0 at  = 0.05 H 0 : π = 0.08 H 1 : π  0.08 Critical Values: ± 1.96 Test Statistic: Decision: Conclusion: z 0 Reject.025 1.96 -2.47 There is sufficient evidence to reject the company’s claim of 8% response rate. -1.96

43. Do not reject H 0 Reject H 0  /2 =.025 1.96 0 Z = -2.47 Calculate the p-value and compare to  (For a two-tail test the p-value is always two-tail) (continued) p-value = 0.0136: p-Value Solution Reject H 0 since p-value = 0.0136 <  = 0.05 Z = 2.47 -1.96  /2 =.025 0.0068

44. 44 Example 1  Price/earnings ratios for stocks. Theory: stable rate of P/E in market = 13.  If P/E (market) < 13, you should invest in the stock market.  If P/E (market) > 13, you should take your money out.  We have a sample of 50  = 12.1. Historical σ = 3.0456  Can we estimate if the population P/E is 13 or not?

45. 45 Estimate Steps  Common steps: Set hypothesis: H 0 : μ = 13 H A : μ ≠ 13 Select  =.05.

46. 46 Calculating Test Statistic

47. 47 p-value Approach  Calculate the p-value. We will calculate for the lower tail → then make an adjustment for the upper tail.

48. 48 p-value, Two-Tailed Test 0 Z=2.09z Z=–2.09 p(z < –2.09) = ?? p(z > 2.09) = ?? We can just calculate one value, and double it.

49. 49 Calculating the p-value cont’d  Find 2.090 on the Standard Normal Distribution tables: z.00.01.02.03.04.05.06.07.08.09 … 1.9 2.0.4772.4778.4783.4788.4793.4798.4803.4808.4812.4817 2.1 …

50. 50 p-value, Two-Tailed Test 0 Z=2.09z Z=–2.09 p(z < –2.09) = ?? p(z > 2.09) = 0.5 -.4817 = 0.0183 Doubling the value, we find the p-value = 0.0366 0.4817

51. 51 Should We Reject the Null Hypothesis?  Yes! p-value = 0.0366 <  = 0.05.  There is only a 3.66% chance that the measured price/earnings ratio sample mean of 12.1 is not equal to the stable rate of 13 by random chance.

52. Example #2: You want to test the hypothesis that a treatment conducted at the ward you are working on in a hospital is more beneficial than the average treatment used in other wards of the hospital.Given that the mean wellness score of the people on your ward is 87 (N=20), and that the mean for the entire hospital is 76 and the standard deviation of population is 15, is your ward significantly better?  State H o and H 1  State if you’re using a one- or two-tailed test and why.  State the p-value that supports your claim.

53. Hypothesis Testing w/ One Sample Smaller Portion = p <.0006

54. Example A family therapist states that parents talk to their teenagers an average of 27 minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time parents spend in conversation with their teenage children. For the n = 12 parents, the study revealed that following times (in minutes) devoted to conversation in a week:

55. Example Do the psychologists findings differ significantly from the therapist’s claim? If so, is the family expert’s claim an overestimate or underestimate of actual time spent talking to children? Use the 0.05 level of significance with two tails. Mean = 24.58, s2 = 12.24, (s/√n) = 1.00 292219252728 212224263022

56. Hypothesis Testing w/ One Sample  Example #2: H 0 = μ = 27 Critical t (df = 11) = ±2.201 Reject H 0, and conclude that the data are significantly different from the therapist’s claim

57. 57 J) Hypothesis Tests, 2 Means:  ’s Unknown  Two datasets –> is the mean value of one larger than the other? Is it larger by a specific amount?  μ 1 vs. μ 2 –> μ 1 – μ 2 vs. D 0.  Often set D 0 = 0 –> is μ 1 = μ 2 ?

58. 58 Example: Female vs. Male Salaries  Saskatchewan 2001 Census data: - only Bachelor’s degrees - aged 21-64 - work full-time - not in school  Men:  M = $46,452.48, s M = 36,260.1, n M = 557.  Women:  W = $35,121.94, s W = 20,571.3, n W = 534.   M –  W = $11,330.44 } our point estimate.  Is this an artifact of the sample, or do men make significantly more than women?

59. 59 Hypothesis, Significance Level, Test Statistic  Research hypothesis: men get paid more: 1. H 0 : μ M – μ W 0 2. Select  = 0.05 3. Compute test t-statistic:

60. 60 4. a. Compute the Degrees of Freedom  Can compute by hand, or get from Excel:  Critical value of t=1.65 = 888

61. 61 4. b. Computing the p-value Degrees of Freedom 0.200.100.020.0250.010.005 … 100.8451.2901.6601.9842.3642.626 .8421.2821.6451.9602.3262.576 6.375 up here somewhere The p-value <<< 0.005.

62. 62 5. Check the Hypothesis  Since the p-value is <<< 0.05, we reject H 0.  We conclude that we can accept the alternative hypothesis that men get paid more than women at a very high level of confidence (greater than 99%).

63. 63 Excel t-Test: Two-Sample Assuming Unequal Variances MaleFemale Mean46452.4793535121.94195 Variance1314794928423178351.3 Observations557534 Hypothesized Mean Diff.0 df888 t Stat6.381034789 P(T<=t) one-tail1.41441E-10.00000000014 t Critical one-tail1.646571945 P(T<=t) two-tail2.82883E-10 t Critical two-tail1.962639544