Student’s t-distribution

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

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

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 - ∞ - +∞

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

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

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

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

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

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.

t t t =2.262 ANSWER Find the t-value 1-α/2 1-0.05/2 0.975 1-α/2 1-0.05/2 0.975 df=n-1 df=10-1 df=9

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

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

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

EXERCISE In a cancer institute a research was done on a drug to prolong life in patients with throat Ca . 200 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 = 4.6 month S= 2.5 month Placebo X= 3.4 month S= 3.1 month

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

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 =---------------------------------------- 100+100-2 Sp =2.816

Find the t-value t t t = 1.9719 1-α/2 1-0.05/2 0.975 1-α/2 1-0.05/2 0.975 df=n1+n2-2 df=100+100-2 df=198

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 {(4.6-3.4) ±(1.9719)(2.816) √2/100} 95% CI {0.415-1.985 }

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

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)

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

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.

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

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

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

t t t = 2.201 Find the t-value 1-α/2 1-0.05/2 0.975 1-α/2 1-0.05/2 0.975 df=n-1 df=12-1 df=11

_ 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 {0.66-1.28 }

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.

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

Exercise The serum complement activity in 2 samples of diseased and healthy subjects revealed the following : _ no. X S ------------------------------------------------- Diseased 10 62.6 33.8 Healthy 20 47.2 10.1 Assuming normality, find the 95% CI for the difference in means. Do not assume equal variances

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

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