biostatist

Student’s t-distributionWGA

Group 1
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
Patient’s No. 37.6
39.1
1
37.8
39.6
2
37.9
38.8
3
38.4
39.4
4
37.7
38.4
5
37.9
38.2
6
38.3
39.2
7
37.8
39.5
8
38.2
39.3
9
38.4
39.1
10
38.5
38.8
11
37.9
38.6
12



d2
d
After
Before
Patient No.
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
38.4
5
0.09
0.3
37.9
38.2
6
0.81
0.9
38.3
39.2
7
2.89
1.7
37.8
39.5
8
1.21
1.1
38.2
39.3
9
0.49
0.7
38.4
39.1
10
0.09
0.3
38.5
38.8
11
0.49
0.7
37.9
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 = 0.49Find the t-value t t t = 2.201 1-α/2 1-0.05/2 0.975 df=n-1 df=12-1 df=11Apply the formula _ CI {d± t (1-α/2, df=n-1) Sd/√n}=1- α95 % CI {0.97± (2.201)(0.49/√12 ) }95 % CI {0.66-1.28 }

Group 3

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.

ANSWER

Find the t-valuet t t =2.262 1-α/2 1-0.05/2 0.975 df=n-1 df=10-1 df=9 _95 % CI{ X ± t 1-α/2, df=n-1 x SE} 95 % CI {35.7 ± 2.262 x 0.7/√10 }95 % CI {35.15-36.17 }

Group 4

A breast CA research team collected the following data on tumor size: _ Type of tumor n X S ----------------------------------------------------------- A 21 3.85cm 1.95cm B 16 2.80cm 1.70cm Construct 95% CI for the difference between populations means, assuming unequal population variances.

Answer

(S12/n1 + S22/n2)2 df=--------------------------------- (S12/n1)2/n1 + (S22/n2)2/n2=36.4 =36 =35 Find the t-valuet t t = 2.03011-α/2 1-0.05/2 0.975df=n1+n2-2 df=21+16-2 df=35

Answer

_ _CI{(X1-X2) ± t √S12/n1 + S22/n2} 1-α/2 df=n1+n2-2CI{(3.85-2.8) ±2.0301√(1.95)2/21 +(1.7)2/16 }95%CI{1.05 ±1.22 }95% CI{- 0.17 _ 1.22 }


Group 5
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: X S Drug 4.6 month 2.5 month Placebo 3.4 month 3.1 month Assuming normality , find the 95% CI for the difference in means , if the variances are assumed to be equal.

Find the pooled SD

(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 = 1.97191-α/2 1-0.05/2 0.975 df=n1+n2-2 df=100+100-2 df=198 _ _CI {(X1-X2) ± t 1-α/2, df=n1+n2-2 Sp√1/n1 +1/n2}95 % CI {(4.6-3.4) ±(1.9719)(2.816) √2/100}95 % CI {0.415-1.985 }