Download presentation

Presentation is loading. Please wait.

Published byCandace Crosthwaite Modified over 2 years ago

1
Example: In a heart study the systolic blood pressure was measured for 24 men aged 25 and for 30 men aged 40. Do these data show sufficient evidence to conclude that the older men have a higher systolic blood pressure, at the 0.05 level of significance? Since The variable concerning systolic blood pressure is continuous The sample size of each group is greater than 10 Systolic blood pressure values in each group is normally distributed There are two groups and they are independent Independent samples t-test is used

3
24122,833316,7790 30133,666717,3013 GROUP 20- year-old 40- year-old NMeanStd. Deviation 3024N = GROUP 40- year-old20- year-old Mean 1 SD SBP 160 150 140 130 120 110 100

4
(1) H 0 : 1 = 2 H a : 1 < 2 (2) Testing the equality of variances Accept H 0. Variances are equal. H 0 : 2 1 = 2 2 H a : 2 1 2 2

5
(3) (4) t (52,0.05) =1.65< p<0.05, Reject H 0. (5) The older men have higher systolic blood pressure

6
Example: NK cell activity was measured for three groups of subjects: those who had low, medium, and high scores on Social Readjustment Rating Scale. The original observations, sample sizes, means and standard deviations are given in table. Is the mean NK cell activity different in three groups?

8
The variable concerning NK cell activity measures is continuous. There are three groups. Variances are equal in three groups. NK cell activity values in each group are normally distributed. The sample size of each group is greater than 10 One-Way ANOVA

9
12 13 N = GROUP HighModerateLow Mean 1 SD NKA 60 50 40 30 20 10 0

10
H 0 : 1 = 2 = 3 H a : Not all the i are equal. SSA=SST-SSW=7182.31-3394.01=3788.30

11
MSW=SSW/34=3394.01/27=99.82 MSA=SSA/(3-1)=3788.30/2=1894.15 F=MSA/MSW=1894.15/99.82=18.98 We conclude that not all population means are equal.

12
Since n 1 n 2, reject H 0 if 2.46<8.28, accept H 0. H 0 : 2 = 3 19.87>8.12, reject H 0. H 0 : 1 = 3 22.32>8.12, reject H 0. H 0 : 1 = 2 Statistical Decision LSDHypothesis

14
Example: A study was conducted to see if a new therapeutic procedure is more effective than the standard treatment in improving the digital dexterity of certain handicapped persons. Twenty-four pairs of twins were used in the study, one of the twins was randomly assigned to receive the new treatment, while the other received the standard therapy. At the end of the experimental period each individual was given a digital dexterity test with scores as follows.

15
Since The variable concerning digital dexterity test scores is continuous The sample size is greater than 10 digital dexterity test score is normally distributed There are two groups and they are dependent Paired sample t-test

16
NewStandardDifference 4954-5 564214 70637 83776 83 0 685117 84822 63549 67625 79718 88826 4850-2 524111 73676 5257-5 73703 78726 64622 71647 4244-2 51447 564214 40355 81738 Total129 Mean65,4660,085,38 SD14,3814,465,65 H 0 : d = 0 H a : d > 0 t (23,0.05) =1.7139 We conclude that the new treatment is effective. Since, reject H 0.

17
Example: We want to know if children in two geographic areas differ with respect to the proportion who are anemic. A sample of one-year-old children seen in a certain group of county health departments during a year was selected from each of the geographic areas composing the departments’ clientele. The followig information regarding anemia was revealed. Geographic Area Number in sample Number anemic Proportion 14501050.23 23751200.32

18
Reject H 0. We concluded that the proportion of anemia is different in two geographic areas.

19
Example: A special diet program was given to 20 clinically obese people. Subjects’ BMI were measured before the diet and they have been followed for two months. BMI measures before the diet and after the end of each month following the diet are given in the table. Is the diet program effective?

20
The variable concerning BMI is continuous The sample size is greater than 10 BMI values are normally distributed There are three groups and they are dependent Repeated measures ANOVA

22
Sources of variationSSdfMSFSig. Times16.79 2 8.3995.920.000 Subjects567.721929.89 Error3.33380.09 Diet program is effective on obese subjects’ BMI.

23
Example: To test the median level of energy intake of 2 year old children as 1280 kcal reported in another study, energy intakes of 10 children are calculated. Energy intakes of 10 children are as follows: Child12345678910 Energy Intake 15008251300170097012001110127014601090

24
Since The variable concerning energy intake is continuous The sample size is not greater than 10 Energy intake is not normally distributed There is only one group Sign test

25
H 0 : The population median is 1280. H A : The population median is not 1280. Child12345678910 Energy İntake 15008251300170097012001110127014601090 ++--+----+ Number of (-) signs = 6 and number of (+) signs = 4 For k=4 and n=10 From the sign test table p=0.377

26
Since p > 0.05 we accept H 0 We conclude that the median energy intake level in 2 year old children is 1280 kcal.

27
Example: Cryosurgery is a commonly used therapy for treatment of cervical intraepithelial neoplasia (CIN). The procedure is associated with pain and uterine cramping. Within 10 min of completing the cryosurgical procedure, the intensity of pain and cramping were assessed on a 100-mm visual analog scale (VAS), in which 0 represent no pain or cramping and 100 represent the most severe pain and cramping. The purpose of study was to compare the perceptions of both pain and cramping in women undergoing the procedure with and without paracervical block.

28
5 women were selected randomly in each groups and their scores are as follows: GroupScore Women without a block 14 88 37 27 0 Women with a paracervical block 50 70 37 66 75

29
Since The variable concerning pain/cramping score is continuous The sample size is less than 10 There are two groups and they are independent Mann Whitney U test

30
GroupScoreRank I01 I142 I273 I374.5 II374.5 II506 II667 II708 II759 I8810 R1= 1+2+3+4.5+10 = 20.5 From the table, critical value is 21 19.5 < 21 accept H 0 We conclude that the median pain/ cramping scores are same in two groups.

31
Example: A study was conducted to analyze the relation between coronary heart disease (CHD) and cigarette smoking. 40 patients with CHD and 50 control subjects were randomly selected from the records and smoking habits of these subjects were examined. Observed values are as follows: + - Yes No Total 90 Smoking Total CHD 30 4 46 1476 40 50 10

32
Observed and expected frequencies + - Yes No Total 90 CHD Total Cigarette Smoking 30 4 46 1476 40 50 10 6.2 33.8 7.842.2

33
df = (r-1)(c-1)=(2-1)(2-1)=1 2 (1,0.05) =3.845 Conclusion: There is a relation between CHD and cigarette smoking. 2 =4. 95 > reject H 0

34
Example:To test whether the weight-reducing diet is effective 9 persons were selected. These persons stayed on a diet for two months and their weights were measured before and after diet. The following are the weights in kg: Subject Weights BeforeAfter 18582 29192 36862 47673 58281 68783 710585 89388 99890 Since The variable concerning weight is continous. The sample size is less than 10 There are two groups and they are dependent Wicoxon signed ranks test

35
Subject WeightsDifference D i Sorted D i Rank Signed Rank BeforeAfter 1858231.5-1.5 2919211.5 36862633.5 47673333.5 582811455 687834566 71058520677 893885888 9989082099

36
T = 1.5 reject H 0, p<0.05 T = 1.5 < T (n=9, =0.05) = 6 T = 1.5 < T (n=9, =0.01) = 2 reject H 0, p<0.01 We conclude 99% cinfident that diet is effective.

37
Example: Hamilton depression scores was measured for three groups of subjects and shown in table. Is the Hamilton depression scores different in three groups? SubjectGroup1Group2Group3 112015 2111915 312019 413174 59141 612119 722021 811518

38
Since The variable concerning Hamilton depression score is continuous The sample size is less than 10 There are three groups and they are independent Kruskal Wallis H 0 : The population distributions are all identical. H a : At least one of the populations tends to exhibit larger values than at least one of the other populations.

39
Group1Group2Group3 ScoreRankScoreRankScoreRank 13.52021.51514 11919 1514 12100119 1311171647 98141213.5 1 2123.519 262021.52123.5 13.515141817 RiRi 54.5128.5117

40
KW=7.93 > Reject H 0 We conclude that at least one of the populations tends to exhibit larger values than at least one of the other populations.

41
Multiple Comparisons Table GroupsStatistical Decision 1-29.255.87p<0.05 1-37.825.87p<0.05 2-31.435.87P>0.05

42
Example: To compare the effects on the clotting time of plasma of four different methods of treatment of the plasma. Samples of plasma from 8 subjects were assigned to the four treatments. Treatments SubjectIIIIIIIV 18.49.49.812.1 212.815.212.914.4 39.69.111.29.8 4 8.89.912.0 58.48.28.5 68.69.99.810.9 78.99.09.210.4 87.98.18.210

43
Since The variable concerning clotting time is continuous The sample size is less than 10 There are four groups and they are dependent Friedman Test

44
Treatments IIIIIIIV timeranktimeranktimeranktimerank 8.419.429.8312.14 12.8115.2412.9214.43 9.629.1111.249.83 28.819.9312.04 8.428.218.53.58.53.5 8.619.939.8210.94 8.919.029.2310.44 7.918.128.23104 Sum R i 111623.529.5

45
2 (3,0.05) = 7.815 < F r =14.96, p<0.05 Reject H 0. We conclude that at least one of the treatments were different from the other treatments.

46
Measures of lipid content and total energy content of the stools were recorded. Data for the cystic fibrosis children are given in the table. Find and interpret the correlation between stool lipid and stool energy Example

48
p=0.06251 There is no significant correlation between stool lipid and stool energy. Correlation between stool lipid and stool energy is 0.42. Is it significant?

49
Example Intravenous glucose tolerance tests were performed on 14 hyperthyroid women, 6 of whom were overweight, and in 19 volunteers with normal thyroid levels matched for age and weight. Insulin sensitivity and BMI of these women are given in the table. Find the correlation coefficients between BMI and insulin sensitivity in the control group and hyperthyroid group. In which group, bmi is more effective? Find the regression equation for each group?

51
BMI_0:INSULIN_0: r2 = 0,2083; r = -0,4564, p = 0,0495; y = 1,32639193 - 0,0299838014*x BMI_0:INSULIN_0: r2 = 0,6006; r = -0,7750, p = 0,0011; y = 2,33571397 - 0,0767480919*x

53
In the hyperthyroid group, 60.1% of the variation in insulin sensitivity is explained by BMI. Rest of the variation is explained by another independent variable(s).

54
In hyperthyroid group, each time a woman’s BMI increases by 1 unit, her predicted insulin sensitivity decreases by 0.077. For example, as the BMI increases from 25 to 30, insulin sensitivity decreases from about 0.41 to about 0.03.

55
Predictive values of insulin sensitivity in hyperthyroid group.

Similar presentations

Presentation is loading. Please wait....

OK

Inference for Distributions

Inference for Distributions

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on types of abortion Ppt on network switching systems Ppt on seasons for kindergarten Ppt on cloud computing services Ppt online mobile shopping Ppt on statistics in maths what is pi Ppt on total quality management in education Ppt on collective nouns for grade 3 Ppt on sudoku game project Ppt on db2 architecture and process