Download presentation

Presentation is loading. Please wait.

Published byKennedy Pickering Modified over 3 years ago

1
- ONE SAMPLE HYPOTHESIS TESTS - TWO SAMPLE HYPOTHESIS TESTS (INDEPENDENT AND DEPENDENT SAMPLES) 1 June 7, 2012

2
2 Example: A researcher believes that mean hemoglobin value is 12 in the population. He selected 64 adults from population randomly to verify this idea and identified hemoglobin mean as 11.2. According to these findings, is population hemoglobin mean different from 12? Since The variable hemoglobin is continuous The sample size is 64 Hemoglobin is normally distributed There is only one group One sample t test is used

3
3 NMeanStd. Dev. Hemoglobin6411.22.8 Hypothesis: H 0 : = 12 H a : 12 Sample results: Population mean: 12 Test statistic:Since the population variance is unknown, our test statistic is t Level of significance: =0.05

4
4 t ( /2,n-1) = t (0.025,64-1) 2.00 t cal > t table Reject H 0 Population mean is different from 12.

5
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 5

6
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 6

7
H 0 : The population median is 1280. H A : The population median is not 1280. Child 12345678910 Energy intake Sign 15008251300170097012001110127014601090 ++--+----+ Number of (-) signs = 6 and number of (+) signs = 4 For k=4 and n=10 From the sign test table p=0.377 7

8
Since p > 0.05 we accept H 0 We conclude that the median energy intake level in 2 year old children is 1280 kcal. 8

9
9 Example: The dean of the faculty wants to know whether the smoking ratio among the Phase I students is 0.25 or not. For this purpose, 50 student is selected and 18 of them said they were smoking. According to this results, can we say that this ratio is different from 0.25 at the 0.05 level of significance? H 0 : P=0.25 H A : P 0.25 Critical z values are 1.96 1.80<1.96 Accept H 0. Smoking ratio among the Phase I student is 0.25. p=18/50=0.36 One sample z test for proportion / one sample chi-square test

10
10 We can solve this problem with one sample chi square test at the same time. SmokingObservedExpected(O-E)(O-E) 2 (O-E) 2 /E Yes1812.5 (50*0.25) 5.530.252.42 No3237.5 (50*0.75) -5.530.250.81 Total50 03.23

11
11 SPSS Output

12
Example: In a heart study the systolic blood pressure was measured for 24 men aged 20 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 12

13
13

14
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 14

15
(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 15

16
(3) (4) t (52,0.05) =1.675 < p<0.05, Reject H 0. (5) The older men have higher systolic blood pressure 16

17
17 SPSS Output

18
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. 18

19
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 19

20
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 20

21
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. 21

22
22 SPSS Output

23
23

24
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 24 The difference between two population proportion

25
Reject H 0 We concluded that the proportion of anemia is different in two geographic areas. 25

26
Example: A study was conducted to analyze the relation between coronary heart disease (CHD) and 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 26

27
Observed and expected frequencies + - Yes No Total 90 Smoking Total CHD 30 4 46 1476 40 50 10 6.2 33.8 7.8 42.2 27

28
df = (r-1)(c-1)=(2-1)(2-1)=1 2 (1,0.05) =3.841 Conclusion: There is a relation between CHD and smoking. 2 =4. 95 > reject H 0 28

29
29 SPSS Output

30
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. 30

31
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 31

32
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) =1.714 We conclude that the new treatment is effective. Since, reject H 0. 32

33
33 SPSS Output

34
34

35
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 Wilcoxon signed ranks test 35

36
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

37
T = 1.5 reject H 0, p<0.05 T = 1.5 < T (n=9,a =0.05) = 6 We conclude that the diet is effective. 37

38
38

39
39 Example: 35 patients were evaluated for arrhythmia with two different medical devices. Is there any statistically significant difference between the diagnose of two devices? Device I Device II Total Arrhythmia (+)Arrhythmia (-) Arrhythmia (+)10313 Arrhythmia (-)13922 Total231235 The significance test for the difference between two dependent population / McNemar test

40
40 H 0 : P 1 =P 2 H a : P 1 P 2 Critical z value is ±1.96 Reject H 0

41
41 McNemar test approach: 2 (1,0.05) =3.841<5.1 p<0.05; reject H 0.

42
42 SPSS Output

Similar presentations

OK

1 Testing Statistical Hypothesis Independent Sample t-Test Heibatollah Baghi, and Mastee Badii.

1 Testing Statistical Hypothesis Independent Sample t-Test Heibatollah Baghi, and Mastee Badii.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on ectopic pregnancy Ppt on reliance securities limited Ppt on p&g products promotion Ppt on mahatma gandhi movements Ppt on p&g products brands tide Ppt on carburetor working animation Ppt on atrial septal defect in adults Ppt on chromosomes and genes and alleles Ppt on staff motivation Ppt on depth first search youtube