T test for two Independent Samples. Raja, BSc.N, DCHN, RN Nursing Instructor Acknowledgement: Ms. Saima Hirani June 07, 2016

Transcription

1 T test for two Independent Samples Raja, BSc.N, DCHN, RN Nursing Instructor Acknowledgement: Ms. Saima Hirani June 07, 2016

2 Q1. The mean serum creatinine level is measured in 36 patients after they received a newly proposed antibiotic was 1.2 mg/dl. If the mean and standard deviation of serum creatinine in the general population are 1.0 & 0.4 mg/dl respectively. Using a significance level of 0.05, test if the mean serum creatinine level in this group is different from that of the general population using the above information.

3 Z test

4 Q2. The mean serum creatinine level is measured in 25 patients after they received a newly proposed antibiotic was 1.1 mg/dl and the sample standard deviation is 0.6 mg/dl. If the mean of serum creatinine in the general population is 0.9 mg/dl. Using a significance level of 0.05, test if the mean creatinine level in this group is different from that of the general population using the above information. serum

5 t test

6 Its similar to the standard normal distribution in some ways like: It is bell shaped and symmetrical about the mean The mean, median, and mode are equal to 0 and located at the center of the distribution Area under the curve is equal to 1

7 When is unknown then the t statistic uses estimated standard error from the sample, not population standard error t x s n

8 x t, d f 2 s n d.f. = n-1

9 Q3. The mean serum creatinine level is measured in 13 white women who received a newly proposed antibiotic was 1.2 mg/dl and the standard deviation was 0.6 mg/ dl. Another sample of 12 black women who have received an old antibiotic have mean serum creatinine level of 1.0 mg/dl with standard deviation of 0.4 mg/dl. Using a significance level of 0.05, test if the mean serum creatinine level in white women is different than the black women.

10 Two groups Two means Two SD Comparison of two means By t- test

11 Decide 2 samples are independent or dependent on each other. Independent: If there is no connection between two samples then Perform t- test for two independent samples

12 Compares means of two groups Experimental treatment versus control Existing groups males versus females We often collect sample from two independent normal population With different means 12

13 1. The samples have been randomly selected of size n1 & n2 from two independent populations 2. The underlying populations are normally distributed 3. Population variance assumed, such that 1 2 may be known or = 2 2 = 2

14 Conclusion 5 Critical region/ P value 4 Test Statistics 3 2 Level of significance = α 1 State null and Alternative hypotheses t TEST 14

15 Group I n n1 s1 Group II n n2 s2 X1 X2

16 STEP I: Lower tail test: H 0 : μ 1 μ 2 H 1 : μ 1 < μ 2 Upper tail test: H 0 : μ1 μ2 H 1: μ 1 > μ2 Two-tailed test: H 0 : μ 1 = μ 2 H 1 : μ 1 μ 2

17 STEP II: Level of Significance α = STEP III: Test Statistic =

18 Now since 2 is unknown, we need to find an estimate for it. Using the variances from the two samples S 2 1 & S 2 2 we obtain an improved estimate for 2 called the pooled sample variance, Sp 2 n 1 s 2 n 1 s 2 s 2 p n 1 n 2 2

19 STEP IV: Critical region: for significance level α and df = n1+ n2-2 rejection rules: Reject H 0 if t cal > t tab or t cal <- t tab STEPV: Conclusion

20 Suppose an investigator wants to evaluate the impact of health education on mothers registered to get antenatal care. He asked 25 newly registering mothers, pregnant for the first time, who have not received health education to fill out a questionnaire which tests their knowledge about pregnancy. He wants to compare the knowledge of these mothers, to those who have had health education. He selected the second sample from mothers, who are pregnant for the first time, who are registered with our hospital, but who have attended health education classes about pregnancy. He gave them the same questionnaire. This is his data: 20

21 No Health Education Health Education Number of mothers (n) 25 ( n 1 ) 27 ( n 2 ) Mean score ( X) 12.5 ( X 1 ) 16.4 ( X 2 ) Standard Deviation (S) 4.5 ( S 1 ) 4.2 ( S 2 ) Run a hypothesis test to identify is there a difference in the knowledge about pregnancy between these two groups? 21

22 H 0 : 1 = 2 H a : 1 : 2 = 0.05 t = ( X1 - X2). Sp2 (1/ n1 + 1/ n2 ) Sp 2 = (n1-1) S1 2 + (n2-1) S2 2 n1 + n2-2 Sp 2 = (25-1) (4.5) 2 + (27-1) (4.2) 2 = t = ( ) (1/25 + 1/27)

23 Cont Rejection Region : Reject Ho if t cal > t tab or t cal <- t tab As we know Degrees of freedom = 50, = 0.05, /2= So T tab = 2.01 Hence t cal < Since t cal (-3. 23) is greater than t tab (-2.01) and falls in critical region so we reject null hypothesis at 5% level of significance in favor of Ha and we have sufficient evidence to conclude that the knowledge about pregnancy differs for the two groups 23

24 24

25 A researcher wishes to determine whether the salaries of professional nurses employed by private hospitals are higher than those of nurses employed by government hospitals. She selects a sample of nurses from each type of hospital and calculates the means and sd of their salaries. At α = 0.01, can she conclude that the private hospitals pay more than the government hospitals? Assume that the populations are approximately normally distributed.

26 Data: Private X = $ 26,800 s1 = $ 600 n1 = 10 Government X= $ 25, 400 s2 = $ 450 n2 = 8

27 Step I: Ho : µ1 µ2 Ha: µ1 > µ2 Step II: α = 0.01

28 Step III: t statistic = s p n 1 s 2 n 1 s n n 2 1 2

29 Step IV: Critical region: t tabα, d.f. = t tab, 0.01, 16 =2.583 Reject Ho if t cal> t tab As t cal 5.47 > t tab 2.583

30 Step V: Since our t cal (5.47)is greater than t tab (2.583) and falls under critical region, so we reject our Ho at 1 % level of significance in favor of Ha and we have sufficient evidence to conclude that the salaries paid to nurses employed by private hospitals are higher than those paid to nurses employed by government hospitals.

31 ( x x ) t 1 2 2,d f 1 s 2 p n 1 1 n 2