Chapter 4(2) Hypothesisi Testing
- 1. Hypothesis Testing for Continuous Variables MBBS.WEEBLY.COM Chapter4
- 2. We have learned: Basic logic of hypothesis testing Main steps of hypothesis testing Two kinds of errors One-side & two-side test One-sample t test
- 3. 4.3 The t Test for Data under Randomized Paired Design
- 4. 4.3 The t Test for Data under Randomized Paired Design Example 4.2 The weights (kg) of 12 volunteers were measured before and after a course of treatment with a “new drug” for losing weight. The data is given in Table 4.1. Please evaluate the effectiveness of this drug.
- 6. Solution:
- 7. P ＞ 0.50
- 8. 4.4 The Tests for Comparing Two Means Based on Two Groups of Data under Completely Randomized Design
- 9. 4.4 The t test for Comparing Two Means About this design: The individuals are randomly divided into two groups which correspond to two treatments respectively; The two groups are randomly selected from two populations respectively.
- 10. Example 4.3 Assume the red cell counts of healthy male residents and healthy female residents of certain city follow two normal distributions respectively, of which the population means are different and population standard deviations are equal.
- 11. There are two random samples drawn from the two populations respectively, of which the sample sizes, sample means and sample standard deviations are ; ; . It is expected to estimate the 95% confidence interval for the difference of average red cell counts between healthy male and female residents .
- 14. Question: Please judge whether the population means of males and females are equal or not.
- 15. 4.4.1 Equal Variances
- 16. Solution: P ＜ 0.01
- 17. 4.4.2 Unequal Variances Satterthwaite’s method
- 18. Example 4.4 n 1 =10 patients and n 2 =20 healthy people are randomly selected and measured for a biochemical index. The mean and standard deviation of the group of patients are =5.05 and S 1 =3.21, and that of the group of healthy people are =2.72 and S 2 =1.52. Please judge whether the two population means are equal or not.
- 19. Solution: P ＞ 0.05
- 20. 4.5 The F -Test for Equal Variances of Two Groups of Data under Completely Randomized Design
- 21. 4.5 The F -Test for Equal Variances of Two Groups
- 22. 1 =n 1 -1, 2 =n 2 -1
- 24. The larger variance is always taken as the numerator of the statistic VR for convenience . Thus, to a two-side test, given , one may use /2 to find the upper critical value F /2 of the F distribution, and if the current value of VR is greater than or equal to F /2 ， then P ≤ ， otherwise, P ＞ ;
- 25. Returning to Example 4.3, where VR =1.09 ， 1 =19 ， 2 =14. Let =0.10 ， the two-side critical value of F distribution is F 0.10/2 = F 0.05 =2.40. Since VR ＜ F 0.05 , not reject , hence the there is no enough evidence to say that the two population variances are not equal .
- 27. 4.6.1 The Z-test for the population probability of binomial distribution ( large n )
- 28. 4.6.1 The Z-test for the population probability n 5 n(1- ) 5
- 29. 4.6.1.1 One sample Example 4.7 150 physicians being randomly selected from the departments of infectious diseases in a city had received a serological test. As a result, 35 out of 150 were positive( 23% ). It was known that the positive rate in the general population of the city was 17%.
- 30. Please judge whether the positive rate among the physicians working for the departments of infectious diseases was higher than that in the general population.
- 31. Solution:
- 33. 4.6.1.2 Two samples Example 4.8 To evaluate the effect of the routine therapy incorporating with psychological therapy, the patients with the same disease in a hospital were randomly divided into two groups receiving routine therapy and routine plus psychological therapy respectively.
- 34. After a period of treatment, evaluating with the same criterion, 48 out of 80 patients (60%) in the group with routine therapy were effective, while 55 out of 75 (73%) in other group were effective. Please judge whether the probability of effective were different in terms of population.
- 35. Solution:
- 38. P =0.08
- 39. 4.6.2 The Z-test for the population mean of Poisson distribution ( largeλ )
- 40. 4.6.2 The Z-test for the population mean of Poisson distribution (largeλ)
- 41. 4.6.2.1 Single observation Example 4.9 The quality control criterion of an instrument specifies that the population mean of radioactivity recorded in a fixed period should not be higher than 50 . Now a monitoring test results in a record of 58 . Please judge whether this instrument is qualified in terms of the population mean .
- 42. Solution:
- 43. P = 0.13
- 44. 4.6.2.2 Two observations Example 4.10 The radioactivity of two specimens was measured for 1 minute independently, resulting in X 1 =150 and X 2 =120 respectively. Please judge whether the two corresponding population means in 1 minute are equal or not.
- 45. Solution:
- 48. 4.6.2.3 Two “ groups ” of observations Example 4.11 The radioactivity of two specimens was independently measured for 10 minutes and 15 minutes respectively, resulting in X 1 =1500 and X 2 =1800 . Please judge whether the two corresponding population means in 1 minute are equal or not.
- 49. Solution:
- 51. THE END THANKS ！