Tests involving Small Sample n < 30Some times it is not practical or even possible to obtain large samples, cost, available time and other factors may require us to work with small samples for our purposes we will say a sample of size less than 30 is a small sample Assumption for t-test 1. 2. 3. The parent population from which sample is drawn is normal The sample is random (small n 2

The test statistic is given by t= x y2 ( n1 1) S12 + ( n2 1) S 2 1 1 ( + ) n1 + n 2 2 n1 n 2

The statistic t has a t-distribution with n1+n2-2 d.f., if it can be assumed that both populations are approximately normal. Example 1 The I.Q of 16 students from one area of city showed a mean of 107 with a standard deviation of 10, while the I.Q.s of 14 students from another area of the city showed a mean of 112 with a standard deviation of 8. Is there a significant difference between the I.Q of the groups? Use (a) .01 and (b) .05 level of significance H 0 : 1 = 2 , there is no difference between the groups H 1 : 1 2 , there is significant difference between the groups Under Ho, the test statistic is t= 112 107 = 1.45 15(100) + 13(64) 1 1 ( n1 1) S + ( n2 1) S 1 1 = ( + ) ( + ) 16 + 14 2 16 14 n1 + n 2 2 n1 n 22 1 2 2

x y

Accept H 0

Reject H 0

Example 2:- At an agricultural station it was desired to test the effect of a given fertilizer on wheat production. To accomplish this, 24 plots of land having equal areas were chosen, half of there were treated with the fertilizer and other half were untreated. Otherwise conditions were same. The mean yield of wheat on the untreated plots was 4.8 kgs. With a standard deviation of 4 kgs, while the mean yield on the treated plots was 5.1 kg. with a standard deviation of 3.6 kg. can we conclude that there is significant improvement in wheat production because of the fertilizer if the significance level of (a) 1 % and (b) 5 % is used. H 0 : 1 = 2 , there is no difference between the groups H 1 : 1 > 2 , there is significant difference between the groups Under Ho, the test statistic is t= 5.1 4.8 = 1.8.5 ( n1 1) S + ( n2 1) S 1 1 1 = 11(16) + 11(3.6) 2 1 ( + ) ( + ) 12 + 12 2 12 12 n1 + n 2 2 n1 n 22 1 2 2

x y

Paired t-test for difference of means(Dependent samples, sample observations are paired) Let us now consider the care when (i) sample sizes are equal n1 = n2 =n (say) and (ii) the two samples are not independent but the sample observations are paired together i.e. the pair of observations ( xi , y i ) corresponds the same i th sample unit. i = 1,2,..., n The problem is to test of the sample means differ significantly. Many statistical problems use paired data samples to draw conclusions about the difference between two populations, data pairs occur may naturally in before and after situations, where the same object or item is measured both before and after a treatment. Example 1:- The diastolic blood pressure of 12 were taken before and after administering a certain drug Blood pressure 94 80 84 86 88 78 91 92 81 90 100 90 before giving drug Blood pressure after giving 89 78 76 87 91 78 93 91 76 90 96 90 to drug Can we say that the drug has reduced the blood pressure In this problem we want to test the claim that drug has reduced the Blood pressure. This means difference between d i = A-B should tend to be positive and the population mean difference values d should also be positive. The null hypothesis that there is no difference among the pairs i.e. d = 0 i.e. H 0 : d = 0 H : d > 0 To test this hypothesis we use the following test statistic d d Sd / n

t=

The statistic t has student t-distribution

Patient No. 1 2 3 4 5 6 7 8 9 10 11 12 S d = 3.09

B.P. before drug A 94 80 84 86 88 78 91 92 81 90 100 96

B.P. after administering the B drug 89 78 76 87 91 78 93 91 76 90 96 90

D= A B 5 2 8 -1 -3 0 -2 1 5 0 4 6

D2 25 4 64 1 9 0 4 1 25 0 16 36

d = 2.583 2.583 0 3.09 / 12

t=

d d Sd / n

=

= 2.89

Reject H 0 and conclude that the drug does reduce the blood pressure Example 2:- An IQ test was administered to 5 persons before and after they were trained. The results are given below:Candidates IQ. Before training IQ. After training I 110 120 II 120 118 III 123 125 IV 132 136 V 125 121

Test Whether any change in I.Q,. after the training programme.

H 0 : d = 0 H : d 0 Candidates I II III IV V S d = 30 IQ. Before training A 110 120 123 132 125 IQ after training B 120 118 125 136 121 d =2 d=BA 10 -2 2 4 -4 D2 100 4 4 16 16

t=

d d Sd / n

=

20 5 6/ 5

= 0.816

We accept H 0 and conclude that there is no effect of training.