1. 1 Outline Small Sample Tests 1. Hypothesis Test for  – Small Samples 2. t-test Example 1 3. t-test Example 2 4. Hypothesis Test for the Population Proportion p – Large Samples 5. Population Proportion Example 1 6. Population Proportion Example 2

2. 2 Hypothesis Testing – Small Samples Small Sample Tests  Central Limit Theorem (review):  For large enough samples, sampling distribution of the mean will be normal even if the raw data are not normally-distributed.  But what do we do when our sample size is not “large enough?”

3. 3 For n < 30, we face two problems: Small Sample Tests a) Central Limit Theorem does not apply  Z does not give probability of finding X in some range relative to μ O, when n < 30.

4. Small Sample Tests When sampling distribution of the mean is normal, the Z table gives us the probability that we will find a sample mean in some range (for samples of size n, with μ = μ 0 ). μ0μ0

5. Small Sample Tests But when n is < 30, we cannot be sure that the sampling distribution of the mean is normal. So, how do we obtain the probability that a sample mean will be in a given range? μ0μ0

6. 6 For n < 30, we face two problems: Small Sample Tests b) s is not a good estimator of .  That is, variability in the sample is not a good source of information about variability in the population.

7. 7 Hypothesis Testing – Small Samples Small Sample Tests  To measure the probability of finding the mean of a small sample in a given range relative to μ O, we use a different probability distribution – the t distribution. t = –  s/√n X

8. 8 Hypothesis test for small samples Small Sample Tests  CAUTION:  The t distribution changes shape with sample size, becoming more like the SND as n gets larger.  For a hypothesis test, to find the critical value of t in the t table, you need to know 2 things: α and degrees of freedom.  For the one-sample t test, d.f. = n – 1.

9. 9 Hypothesis Testing – Small Samples Small Sample Tests  Very important point about testing with n < 30:  If an exam question with n < 30 gives you the population standard deviation, , then use Z.  Large n (≥ 30)use Z  Small n,  knownuse Z  Small n,  unknownuse t

10. 10 Hypothesis Testing – Small Samples Small Sample Tests H 0 :  =  0 H A :  ≤  0 H A :  ≠  0 or H A :  ≥  0 (One-tailed test)(Two-tailed test) Test Statistic:t = -  0 s/√n X

11. 11 Hypothesis Testing – Small Samples Small Sample Tests Rejection Region: One-tailed test:Two-tailed test: t obt > t α │ t obt │ > t α /2 or t obt < -t α  Where t α and t α /2 are based on d.f. = (n – 1)  Remember to report your decision explicitly!!

12. 12 Confidence Interval – Small Samples Small Sample Tests C.I. = ± t α /2 (s/√n)  Notes: t α /2 is based on d.f. = (n – 1). Use this C.I. when n < 30 and  not known. X

13. 13 Example 1 – t-test Small Sample Tests  In a recent pollution report, a team of scientists expressed alarm at the dihydrogen monoxide levels in fish in Ontario lakes. Historically, the average dihydrogen monoxide level has been 2.65 parts per thousand (ppt). This year, samples of fish from 20 lakes in Ontario turned up an average dihydrogen monoxide count of 2.98 ppt with a variance of 36.  Note: for more information on DHMO, visit http://www.dhmo.org/

14. 14 Example 1 – t-test Small Sample Tests a. Does it appear that dihydrogen monoxide levels are increasing in fish in Ontario lakes? ( α =.01) b. The dihydrogen monoxide levels of fish in 5 lakes in the Timmins area were 2.84, 3.96, 4.40, 1.60, and 2.63. Construct a 95% confidence level for the average dihydrogen monoxide counts in fish in the Timmins area.

15. 15 Example 1a – t-test Small Sample Tests H 0 :  = 2.65 H A :  > 2.65 Test Statistic:t = -  0 s/√n Rejection region:t obt > t.01,19 = 2.539 X

16. 16 Example 1a – t-test Small Sample Tests = 2.98s 2 = 36n = 20 t obt = 2.98 – 2.65 √36/20 t obt = 0.246 Decision: Do not reject H 0. There is not enough evidence to conclude that dihydrogen monoxide levels are increasing. X

17. 17 Example 1b – t-test Small Sample Tests Σ X = 15.43 = 15.43/5 = 3.086 Σ X 2 = 52.584( Σ X) 2 = 15.43 2 = 238.085 S 2 = 52.584 – 238.085= 1.242 5 4 S = √1.242 = 1.114S = 1.114/√5 =.498 X X

18. 18 Example 1b – t-test Small Sample Tests  C.I. = ± t α /2 (s/√n)  For 95% C.I., α =.05. Associated t.025,4 = 2.776  C.I. = 3.086 ± 2.776 (.498) = (1.703 ≤  ≤ 4.469) X

19. 19 Example 2 – t-test Small Sample Tests  There have been claims that provincial funding cuts for hospitals have led to an increase in the average waiting time for elective surgery. A search of past records reveals that that average waiting time for elective surgery was 38.5 days prior to the 2003 Ontario election. A random sample of patients scheduled for elective surgery is identified, and the waiting time until surgery for each patient is measured. The data are shown below as # of days intervening between when surgery is ordered and when it occurred.  Is there evidence in these data that average waiting time has increased under the current provincial government? ( α =.01)

20. 20 Example 2 – t-test Small Sample Tests Patient #Waiting time (days) 143 228 355 438 530 645 751 839

21. 21 Example 2 – t-test Small Sample Tests H 0 :  = 38.5 H A :  > 38.5 Test Statistic:t = –  0 s/√n Rejection region:t obt > t.01,7 = 2.998 X Why one-tailed?

22. 22 Example 2 – t-test Small Sample Tests Σ x = 329n = 8 = 41.125 Σ x 2 = 14149 s 2 = 14149 – (329) 2 8 7 s 2 = 88.411 s = √88.411 = 9.402 X

23. 23 Example 2 – t-test Small Sample Tests t obt = 41.125 – 38.5 9.402/√8 t obt = 0.787  Decision: Do not reject H 0. There is not sufficient evidence to conclude that waiting times have increased.  Note: Be sure to give the full decision.

24. 24 Example 3 – t-test Small Sample Tests  It is known that the mean # of errors made on a particular pursuit rotor task is 60.9. A physiologist wishes to know if people who have had a spinal cord injury but who are apparently recovered perform less well on this task. In order to test this, a random sample of 8 people who have had spinal cord injuries is chosen and they are administered the pursuit rotor task. The # of errors each made is: 63, 66, 65, 62, 60, 68, 66, 64

25. 25 Example 3 – t-test Small Sample Tests a. Is there evidence to support the belief that ‘recovered’ patients are impaired in performing this task? ( α =.01) b. Form the 90% C.I. for the mean number of errors committed by recovered patients.

26. 26 Example 3a – t-test Small Sample Tests Note these words: “It is known…”Population information “The mean # of errorsThis is  0. is … 60.9” “In order to test this…”Hypothesis Test!! “a random sample of 8…”n < 30,  not known This calls for a t-test…

27. 27 Example 3a – t-test Small Sample Tests H 0 :  = 60.9 H A :  > 60.9 Test Statistic:t = -  0 s/√n Rejection region:t obt > t.01,7 = 2.998 X Why “greater than?”

28. 28 Example 3a – t-test Small Sample Tests = 64.25s = 2.55n = 8 t obt = 64.25 – 60.9 2.55/√8 t obt = 3.72 Decision: Reject H 0. Recovered patients perform worse than normals on this task. X

29. 29 Example 3b – t-test Small Sample Tests C.I. = ± t α /2 (s/√n) For 90% C.I., α =.10. Corresponding t.05,7 = 1.895. C.I. = 64.25 ± 1.895 (2.55/√8) (62.542 ≤  ≤ 65.958) X