1. Student’s t-distribution

2. Student’s t-distribution
In cases where the population variance σ2 is unknown we can use the sample variance S2 as the best point estimate for the population variance σ2

3. Student’s t-distribution
The distribution will not follow the standard normal distribution (Z distribution), but it will follow the t-distribution

4. Student’s t-distribution
The most important characteristics of t-distribution are :
It has a mean of zero
It is symmetric around the mean
It ranges between - ∞ - +∞

5. Student’s t-distribution
4. Compared to the standard normal distribution the curve is less peaked with higher tails
5. The quantity n-1 which is called the degrees of freedom (df) is used in computing the sample variance
6. The t-distribution approaches the standard normal distribution as the degrees of freedom approaches infinity

6. Student’s t-distribution
The formula for calculating the value of t: _
X-µ
t =
s /√n

7. t-table
t0.995
t0.99
t0.975
t0.95
t 0.90 df 1 2 30 35 50 60
100
120
200

8. CI when population variance is unknown
Confidence Interval for the mean of a normal distribution with unknown population variance , and a small sample size
The reliability coefficient will be the t-value (rather than the Z-value) corresponding to the confidence level, and the degree of freedom

9. Confidence Interval t 1-α/2 df=n-1 CI=Estimator ± R.C x SE
R.C= t-value t
1-α/2
df=n-1

10. EXERCISE
Using specimen obtained from 10 individuals, reveal a mean and standard deviation of calcium content of the teeth of 35.7, and 0.7. Find the 95% CI for the mean calcium content.

11. t t t =2.262 ANSWER Find the t-value 1-α/2 1-0.05/2 0.975
1-α/ /
df=n df= df=9

12. ANSWER 95%CI{ X ± t x SE} 1-α/2 95% CI {35.15-36.17 } _ df=n-1

13. Confidence Interval
CI for the difference between two population means when the population variances are unknown but assumed to be equal
We should first find the Pooled Variance S2p

14. Formula CI=(X1-X2) ± t Sp√1/n1 +1/n2 1-α/2 (n1-1)S12+ (n2-1)S22
_ _
CI=(X1-X2) ± t Sp√1/n1 +1/n2
1-α/2
df=n1+n2-2

15. EXERCISE
In a cancer institute a research was done on a drug to prolong life in patients with throat Ca patients are randomly divided into two equal groups , one kept on the drug and the other on a placebo, the following was found: _
Drug X = month S= 2.5 month
Placebo X= 3.4 month S= 3.1 month

16. Exercise
Assuming normality , find the 95% CI for the difference in means , if the variances are assumed to be equal.

17. Find the pooled SD S2p =------------------------- n1+n2-2
(n1-1)S12+ (n2-1)S22
S2p =
n1+n2-2
(100-1) (2.5)2 + (100-1) (3.1)2 =
Sp =2.816

18. Find the t-value t t t = 1.9719 1-α/2 1-0.05/2 0.975
1-α/ /
df=n1+n df= df=198

19. Answer _ _ CI{(X1-X2) ± t Sp√1/n1 +1/n2 } 1-α/2
_ _
CI{(X1-X2) ± t Sp√1/n1 +1/n2 }
1-α/2
df=n1+n2-2
95%CI {( ) ±(1.9719)(2.816) √2/100}
95% CI { }

20. Pairing CI for the mean difference
Many studies are designed to produce observations in pairs .i.e.: BP, RBS….before and after giving certain treatment (Before, After)
A measurement done by two instruments, individuals, times,….
Every individual here has a pair of readings

21. Pairing Find the difference d _ Find the mean of difference d
Find the standard deviation of the difference Sd
The df= n-1 ( since we have only one sample)

22. Pairing
Apply the formula: _
CI{d± t Sd/√n α/2
df=n-1

23. Exercise
In a pediatrics clinic , a study was carried out to see the effectiveness of a certain antipyretic . Twelve 4-years old girls suffering from influenza had their temperatures measured immediately before and 1- hour after administering the drug . For the following results, find the 95% CI for the mean difference.

24. After
Before
Pat
37.6
39.1 1
37.8
39.6 2
37.9
38.8 3
38.4
39.4 4
37.7 5
38.2 6
38.3
39.2 7
39.5 8
39.3 9 10
38.5 11
38.6 12

25. d2 d
After
Before
Pat
2.25
1.5
37.6
39.1 1
3.24
1.8
37.8
39.6 2
0.81
0.9
37.9
38.8 3
1.00
1.0
38.4
39.4 4
0.49
0.7
37.7 5
0.09
0.3
38.2 6
38.3
39.2 7
2.89
1.7
39.5 8
1.21
1.1
39.3 9 10
38.5 11
38.6 12
13.86
11.6
Total

26. Answer _ ∑d 11.6 d=------- =------- = 0.97 n 12
Sd=√n ∑d2 - (∑d )2/n(n-1)
= √12 (13.86) - ( 11.6 )2 /12 x 11
=0.49

27. t t t = 2.201 Find the t-value 1-α/2 1-0.05/2 0.975
1-α/ /
df=n df= df=11

28. _ CI{d± t Sd/√n}=1- α Apply the formula
1-α/2
df=n-1
95% CI {0.97± (2.201)(0.49/√12 ) }
95% CI { }

29. Exercise
The mean and standard deviation of serum amylase of 15 healthy adult males was 96 units and 35 units /100 ml. Find the 95% CI of the population mean.

30. Exercise
Referring to the previous example , if another sample of 22 hospitalized patients , the mean and standard deviation of serum amylase level was 120, and 40 units/100ml. Assuming normality and equal population variances find the 95% CI for difference in population means

31. Exercise
The serum complement activity in 2 samples of diseased and healthy subjects revealed the following : _
no X S
Diseased
Healthy
Assuming normality, find the 95% CI for the difference in means. Do not assume equal variances

32. Exercise
A cholesterol lowering agent was used on 6 individuals. From the following results find the 95% CI for the mean difference

33. After
(mg/dl)
Before
Patient
209
217 1
241
252 2
230
229 3
208
200 4
206 5
211
213 6