2. 1 Required Sample Size, Type II Error Probabilities Chapter 23 Inference for Means: Part 2

3. Required Sample Size To Estimate a Population Mean  If you desire a C% confidence interval for a population mean  with an accuracy specified by you, how large does the sample size need to be? We will denote the accuracy by ME, which stands for M argin of E rror.

4. Example: Sample Size to Estimate a Population Mean  Suppose we want to estimate the unknown mean height  of male students at NC State with a confidence interval. We want to be 95% confident that our estimate is within.5 inch of  How large does our sample size need to be?

5. Confidence Interval for 

6. Good news: we have an equation Bad news: 1.Need to know s 2.We don’t know n so we don’t know the degrees of freedom to find t * n-1

7. A Way Around this Problem: Approximate by Using the Standard Normal

8. .95 Confidence level Sampling distribution of x

9. Estimating s Previously collected data or prior knowledge of the population If the population is normal or near-normal, then s can be conservatively estimated by s  range 6 99.7% of obs. within 3  of the mean

10. Example: sample size to estimate mean height µ of NCSU undergrad. male students We want to be 95% confident that we are within.5 inch of  so  ME =.5; z*=1.96 Suppose previous data indicates that s is about 2 inches. n= [(1.96)(2)/(.5)] 2 = 61.47 We should sample 62 male students

11. Example: Sample Size to Estimate a Population Mean  -Textbooks Suppose the financial aid office wants to estimate the mean NCSU semester textbook cost  within ME=$25 with 98% confidence. How many students should be sampled? Previous data shows  is about $85.

12. Example: Sample Size to Estimate a Population Mean  -NFL footballs The manufacturer of NFL footballs uses a machine to inflate new footballs The mean inflation pressure is 13.5 psi, but uncontrollable factors cause the pressures of individual footballs to vary from 13.3 psi to 13.7 psi After throwing 6 interceptions in a recent game, Peyton Manning complains that the balls are not properly inflated. The manufacturer wishes to estimate the mean inflation pressure to within.025 psi with a 99% confidence interval. How many footballs should be sampled?

13. Example: Sample Size to Estimate a Population Mean  The manufacturer wishes to estimate the mean inflation pressure to within.025 pound with a 99% confidence interval. How may footballs should be sampled? 99% confidence  z* = 2.576; ME =.025  = ? Inflation pressures range from 13.3 to 13.7 psi So range =13.7 – 13.3 =.4;   range/6 =.4/6 =.067 12348...

14. Significance Levels and Rejections Regions Hypothesis Tests for  13

15. 14 aLevels and Rejection Regions, Right-Tail; n=26 (df=25) If H A :  >  0 and  =.10 then RR={t: t > 1.316} If H A :  >  0 and  =.05 then RR={t: t > 1.708} If H A :  >  0 and  =.01 then RR={t: t > 2.485}  Rej Region.10t > 1.316.05t > 1.708.01t > 2.485

16. 15 Hypothesis Testing for , Type II Error Probabilities (Right-tail example) Example –A new billing system for a department store will be cost- effective only if the mean monthly account is more than $170. –A sample of 401 accounts has a mean of $174 and s = $65. –Can we conclude that the new system will be cost effective?

17. 16 Hypotheses –The population of interest is the credit accounts at the store. –We want to know whether the mean account for all customers is greater than $170. H A :  > 170 –Where  is the mean account value for all customers –We will choose significance level  =.05 Right-tail example: hypotheses, significance level H 0 :  = 170

18. 17 The rejection region: reject H 0 if the test statistic t satisfies t > t.05,n-1 = t.05,400 = 1.649 We will reject H 0 if the value of the test statistic t is greater than 1.649 Results from the n = 401 randomly selected customers: A Right - Tail Test: Rejection Region

19. 18 –Hypotheses: H 0 :  = 170 H A :  > 170 Right-tail example: test statistic and conclusion Recall that the rejection region is Since the test statistic t = 1.23, and 1.23 < 1.649, We do not reject the null hypothesis H 0 :  = 170.

20. 19 P-value: The probability of observing a value of the test statistic as extreme or more extreme then t = 1.23, given that  = 170 is… Right-tail example: P-value and conclusion t 400 Since the P-value >.05, we conclude that there is not sufficient evidence to reject H 0 :  =170. Type II error is possible

21. 20 Calculating , the Probability of a Type II Error Calculating  for the t test is not at all straightforward and is beyond the level of this course –The distribution of the test statistic t is quite complicated when H 0 is false and H A is true – However, we can obtain very good approximate values for  using z (the standard normal) in place of t.

22. 21 Calculating , the Probability of a Type II Error (cont.) We need to 1.specify an appropriate significance level  ; 2.Determine the rejection region in terms of z 3.Then calculate the probability of not being in the rejection when  =  1, where  1 is a value of  that makes H A true.

23. 22 –Test statistic: H 0 :  = 170 H A :  > 170 Choose  =.05 Rejection region in terms of z: z > z.05 = 1.645 Example (cont.) calculating   = 0.05

24. 23 Express the rejection region directly, not in standardized terms  =.05  = 170 Example (cont.) calculating  –The rejection region with  =.05. Do not reject H 0  180 H A :  = 180 H 0 :  = 170 Specify the alternative value under H A. –Let the alternative value be  = 180 (rather than just  >170)

25. 24  =.05  = 170 Example (cont.) calculating   180 H 1 :  = 180 H 0 :  = 170 –A Type II error occurs when a false H 0 is not rejected. Suppose  =180, that is H 0 is false. A false H 0 … …is not rejected

26. 25  = 170 Example (cont.) calculating   180 H 1 :  = 180 H 0 :  = 170 Power when  =180 = 1-  (180)=.9236

27. 26 Increasing the significance level  decreases the value of  and vice versa  Effects on  of changing   = 170  180    2 >  2 <

28. 27 A hypothesis test is effectively defined by the significance level  and by the sample size n. If the probability of a Type II error  is judged to be too large, we can reduce it by –increasing , and/or –increasing the sample size. Judging the Test

29. 28 Increasing the sample size reduces  Judging the Test By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, the cutoff value of for the rejection region decreases.

30. 29  180  = 170 Judging the Test Note what happens when n increases:  does not change, but  becomes smaller Increasing the sample size reduces 

31. 30 Increasing the sample size reduces  In the example, suppose n increases from 400 to 1000. Judging the Test  remains 5%, but the probability of a Type II drops dramatically.

32. 31 A Left - Tail Test Self-Addressed Stamped Envelopes. –The chief financial officer in FedEx believes that including a stamped self-addressed (SSA) envelop in the monthly invoice sent to customers will decrease the amount of time it take for customers to pay their monthly bills. –Currently, customers return their payments in 24 days on the average, with a standard deviation of 6 days. –Stamped self-addressed envelopes are included with the bills for 76 randomly selected customers. The number of days until they return their payment is recorded.

33. 32 The parameter tested is the population mean payment period (  ) for customers who receive self-addressed stamped envelopes with their bill  The hypotheses are: H 0 :  = 24 H 1 :  < 24 Use  =.05; n = 76. A Left - Tail Test: Hypotheses

34. 33 The rejection region: reject H 0 if the test statistic t satisfies t <  t.05,75 =  1.665 We will reject H 0 if the value of the test statistic t is less than  1.665 Results from the 76 randomly selected customers: A Left - Tail Test: Rejection Region

35. 34 The value of the test statistic t is: A Left -Tail Test: Test Statistic Since the test statistic t =  1.52, and  1.52 >  1.665, We do not reject the null hypothesis. Note that the P-value = P( t 75.05. Since our decision is to not reject the null hypothesis, A Type II error is possible. Since the rejection region is

36. 35 The CFO thinks that a decrease of one day in the average payment return time will cover the costs of the envelopes since customer checks can be deposited earlier. What is  (23), the probability of a Type II error when the true mean payment return time  is 23 days? Left-Tail Test: Calculating , the Probability of a Type II Error

37. 36 –Test statistic: H 0 :  = 24 H A :  < 24 Choose  =.05 Rejection region in terms of z: z < -z.05 = -1.645 Left-tail test: calculating  (cont.)  = 0.05

38. 37 Express the rejection region directly, not in standardized terms  =.05  = 23 Left-tail test: calculating  (cont.) –The rejection region with  =.05. Do not reject H 0  24 H A :  = 23 H 0 :  = 24 Specify the alternative value under H A. –Let the alternative value be  = 23 (rather than just  < 24)

39. 38  = 23 Left-tail test: calculating  (cont.)  24 H 1 :  = 23 H 0 :  = 24 Power when  =23 = 1-  (23)=.282  =.05

40. 39 A Two - Tail Test for  The Federal Communications Commission (FCC) wants competition between phone companies. The FCC wants to investigate if AT&T rates differ from their competitor’s rates. According to data from the (FCC) the mean monthly long-distance bills for all AT&T residential customers is $17.09.

41. 40 A Two - Tail Test (cont.) A random sample of 100 AT&T customers is selected and their bills are recalculated using a leading competitor’s rates. The mean and standard deviation of the bills using the competitor’s rates are Can we infer that there is a difference between AT&T’s bills and the competitor’s bills (on the average)?

42. 41 Is the mean different from 17.09? n = 100; use  =.05 H 0 :  = 17.09 A Two - Tail Test (cont.)

43. 42 0  2  0.025 -t   = -1.9842 t   = 1.9842 Rejection region A Two – Tail Test (cont.) t 99

44. 43  2  0.025 -t   = -1.9842 t   = 1.9842 There is insufficient evidence to conclude that there is a difference between the bills of AT&T and the competitor. -1.19 Also, by the P-value approach: The P-value = P(t 1.19) = 2(.1184) =.2368 >.05 1.19 0 A Two – Tail Test: Conclusion A Type II error is possible

45. 44 The FCC would like to detect a decrease of $1.50 in the average competitor’s bill. (17.09-1.50=15.59) What is  (15.59), the probability of a Type II error when the true mean competitor’s bill  is $15.59? Two-Tail Test: Calculating , the Probability of a Type II Error

46. 45  2  0.025 17.09  2  0.025 Rejection region Two – Tail Test: Calculating  (cont.) Do not reject H0 Reject H0

47. 46  = 15.59 Two – Tail Test: Calculating  (cont.)  17.09 H A :  = 15.59 H 0 :  = 17.09 Power when  =15.59 = 1-  (15.59)=.972  =.05

48. General formula: Type II Error Probability  (  A ) for a Level  Test 47

49. Sample Size n for which a level  test also has  (  A ) =  48