1. Testing of Hypothesis

2. Test of Significance A procedure to assess the significance of a statistics that a sample statistics would differ from a given value of the parameter of from another sample value, by more than a certain amount, is known as test of significance. From sampling distribution, it is possible to find the probability that a sample statistics would differ from a parameter or another sample and accordingly a procedure to assess is known as Test.

3. Steps in Testing 1) Lay down the hypothesis Hypothesis Ho: null hypothesis Hα: alternate hypothesis

4. H 0 null hypothesis Null hypothesis is concerned with all possible rejections, while accept one.. In its simplest form, the hypothesis assets that there is no true difference in the sample and the population in the particular matter under consideration. Difference if found, is accidental, unimportant, arising out of fluctuation of sampling. Difference if found, is accidental, unimportant, arising out of fluctuation of sampling.

5. H H α: Alternate Hypothesis AH specifies those values that the researcher believes to hold true and of course, he hopes that the sample states lead to acceptance of this hypothesis as the A AH specifies those values that the researcher believes to hold true and of course, he hopes that the sample states lead to acceptance of this hypothesis as the AH may embrace the whole range of values rather than simple point H α. H 0 µ = 100 H α : µ > 100 Mean IQ = 100 H 0 µ = 50 H α : µ < 100 Mean IQ = 50 µ1 - µ2 = 0 (H 0 ) µ1 - µ2 ≠ 0 (H 0 )

6. II Step: Test Statistics The decision to accept null hypothesis or the alternate H, is made on the basis of a statistic computed from sample. Such a statistic is called the test statistic based on an appropriate probability distribution. III Step : Level of significance The community used levels of significance are 5% (0.05) and 1 % (0.01). 5 % level of 5 samples that in 5 out of 100, we are likely to reject or we are 95% confident that our decision to reject H 0

7. Step IV : Critical Region The critical or the rejection is one, which corresponds to a pre-determined level of significance ‘ The critical or the rejection is one, which corresponds to a pre-determined level of significance ‘α’, whenever sample statistics falls in the critical region, we reject the hypothesis. Step V: Doing computations and making decisions We compute the results from a random sample of size ‘n’ from various tests.

8. Students ‘t” Distribution The word student was the pseudonym of a statistician, named “Gosset”, who developed the theory for small sample in 1905. Applications of t-distribution has a number of applications in statistics. 1. 1.t-test for significance of single mean, population variance 2. t-test for the significance of the difference between two sample means, the population variance being equal but unknown 3. t-test for significance of an observed sample correlation coefficient

9. t-test for Single Mean (  X - µ ) (  X - µ ) t = ----------- n s Where  x = mean of the sample µ = actual or hypothetical; mean of µ = actual or hypothetical; mean of the population the population n = the number of observations n = the number of observations s = standard deviation of the sample s = standard deviation of the sample

10. ∑ (x -  x ) 2 ∑ (x -  x ) 2 S = ----------------- n – 1 n – 1 ∑ d 2 – n ( d ) 2 ∑ d 2 – n ( d ) 2 = ------------------ = ------------------ n – 1 n – 1 Where d= deviation from assumed mean If this calculated value of t exceeds table value t 0.005 We infer that difference between  x and µ is significant at 5 % level.

11. (  x - µ) (  x - µ) t = ------------ n-1 s When we are given sample S.D. we use n-1 in place of n. When we are given sample S.D. we use n-1 in place of n.

12. Example Ten individuals are chosen at random from a population and their heights are found to be 63,63,66,67,68,69,70,70,71 and 71. In the light of the data discuss the suggestion that the mean height in the population is 66 inches.

13. Solution H0: µ = 66 - µ ) (  x - µ ) t = ----------- n t = ----------- n s Height in inches x X - 67 d d2d2 63- 416 63- 416 66- 11 6700 6811 6924 7039 39 71416 71416 ∑ X = 678∑d = 8 ∑d 2 = 88

14. ∑ X 678 ∑ X 678  X = ------- = -------- = 67.8 N 10 N 10 ∑ d 8 ∑ d 8  d = ------- = ------ = 0.8 N 10 N 10 ∑ d 2 – n (  d ) 2 88 - 10 (0.8)2 ∑ d 2 – n (  d ) 2 88 - 10 (0.8)2 S = ------------------ = --------------- = 3.011 S = ------------------ = --------------- = 3.011 n – 1 9 n – 1 9  X = 67.8, S=3.011, µ = 66, n=10  X = 67.8, S=3.011, µ = 66, n=10

15. 67.8 – 66 1.8 67.8 – 66 1.8 t = ------------- 10 = -------- x 3.162 = 1.89 t = ------------- 10 = -------- x 3.162 = 1.89 3.011 3.011 3.011 3.011 V = 10 - 1 = 9 for 9 degrees of freedom the table value of t at 5 % level of significance i.e. t 0.05 = 2.26 V = 10 - 1 = 9 for 9 degrees of freedom the table value of t at 5 % level of significance i.e. t 0.05 = 2.26 Since calculated value of t < t Since calculated value of t < t 0.05 We accept Null Hypothesis Thus we accept that mean height in population is 66 inches.

16. T-test for Difference of Means If we are given two random samples of size n 1 and n 2 with mean  x 1,  x 2 S. D. $ 1 and $ 2. We are interested to know that the two samples come from the same normal population for this we calculate-  x 1 –  x 2 n 1 n 2 t = ------------ -------------- s n 1 + n 2 Where  x 1 = mean of first sample  x 2 = mean of second sample n 1 = number in the first sample n 2 = number in the second sample s = combined standard deviation

17. The value of s is calculated in the following manner: ∑(x 1 -  x 1 ) 2 + ∑ (x 2 -  x 2 ) 2 ∑(x 1 -  x 1 ) 2 + ∑ (x 2 -  x 2 ) 2 s = ------------------------------ n 1 + n 2 – 2 n 1 + n 2 – 2 If we use assumed means then the formula used is : ∑(x 1 – A 1 ) 2 + ∑ (x 2 -A 2 ) 2 –n 1 (x 1 – A 1 ) 2 - n 2 (x 2 -A 2 ) 2 ∑(x 1 – A 1 ) 2 + ∑ (x 2 -A 2 ) 2 –n 1 (x 1 – A 1 ) 2 - n 2 (x 2 -A 2 ) 2 s = --------------------------------------------------------- n 1 + n 2 -2 n 1 + n 2 -2

18. A 1 = assumed mean of the first sample A 2 = assumed mean of second sample x 1 = mean of first sample x 2 = mean of second sample t has n1 + n2 - 2 d.f. If the calculated value of t > t 0.005, the difference between two samples is significant, But if t < t 0.05, the Null hypothesis is accepted. Difference between 2 samples is not significant.

19. Example For a random sample of 10 pigs fed on diet A, the increase in weight in kg. in a certain period were: For a random sample of 10 pigs fed on diet A, the increase in weight in kg. in a certain period were: 10,6,16,17,13,12,8,14,15,9 10,6,16,17,13,12,8,14,15,9 For other random sample of 12 pigs fed on diet B, the increase in the same period were: 7,13,22,15,12,14,18,21,23,10,23,17 For other random sample of 12 pigs fed on diet B, the increase in the same period were: 7,13,22,15,12,14,18,21,23,10,23,17

20. Solution Let us take the Null hypothesis that two diets A and B do not differ significantly. Let us take the Null hypothesis that two diets A and B do not differ significantly.  x 1 –  x 2 n 1 n 2  x 1 –  x 2 n 1 n 2 t = ---------- --------- t = ---------- --------- S n 1 + n 2 S n 1 + n 2 ∑ ( x 1 –  x 1 ) 2 + ∑ ( x 2 –  x 2 ) 2 ∑ ( x 1 –  x 1 ) 2 + ∑ ( x 2 –  x 2 ) 2 s = ------------------------------------- n 1 + n 2 - 2 n 1 + n 2 - 2

21. Increase in Wt. x 1 x 1 -  x 1 (x 1 -  x 1 ) 2 Increase in x 2 X 2 -  x 2 (X 2 -  x 2 ) 2 10-247-864 6-63613-24 164 22749 175251500 131112-39 1200141 8-4161839 142421-749 153923-536 9-3910264 2325 174 ∑x 1 = 1200120∑ x 2 = 1950314

22. ∑ x 1 120 ∑ x 1 120  x 1 = -------- = --------- = 12 n 1 10 n 1 10 195 195  x 2 = ------- = 16.25 12 12

23. ∑ (x 1 –  x 1 ) 2 + ∑ (x 2 –  x 2 ) 2 ∑ (x 1 –  x 1 ) 2 + ∑ (x 2 –  x 2 ) 2 S = ------------------------------- n 1 + n 2 – 2 n 1 + n 2 – 2 120 + 314 434 120 + 314 434 = --------------- = ------- = 4.66 = --------------- = ------- = 4.66 10 + 12 -2 20 10 + 12 -2 20

24. Now  x 1 = 12 x 2 = 15 n 1 = 10 n 2 = 12 s = 4.66 Therefore,  x 1 –  x 2 n 1 n 2  x 1 –  x 2 n 1 n 2 t = ---------- --------- t = ---------- --------- S n 1 + n 2 S n 1 + n 2 12- 16.5 10 x 12 4.5 120 12- 16.5 10 x 12 4.5 120 t = -------- -------------- = --------- ------ 4.66 10 + 12 4.66 22 4.66 10 + 12 4.66 22 4.5 4.5 = -------- x 2.33 = 2.24 = -------- x 2.33 = 2.24 4.66 4.66

25. V = n 1 + n 2 -2 V = n 1 + n 2 -2 V = 10 + 12 -2 = 20 V = 10 + 12 -2 = 20 t 0.05 for v = 20 is 2.09 t 0.05 for v = 20 is 2.09 Calculated value > table value Calculated value > table value We reject Null hypothesis that there is significant difference in diet A and B. We reject Null hypothesis that there is significant difference in diet A and B.

26. Example The mean of a sample of 10 electric bulbs was found to be 1456 hours with S.D. of 423 hours. A second sample of 17 bulbs chosen from a different batch showed a mean life of 1280 hours with S.D. of 398 hours. Is there a significant difference between the means of the two batches?

27. Let the null hypothesis be that two batches do not differ significantly. Let the null hypothesis be that two batches do not differ significantly.  x 1 –  x 2 n 1 n 2  x 1 –  x 2 n 1 n 2 t = ----------- ----------- s n 1 + n 2 s n 1 + n 2 x 1 = 1456 n 1 = 10 s 1 = 423 x 2 = 1280 n 2 = 17 s 2 = 398

28. n 1 s 1 2 +n 2 s 2 2 n 1 s 1 2 +n 2 s 2 2 S = ------------------- n 1 + n 2 -2 n 1 + n 2 -2 10 (423) 2 + 17 (398) 2 10 (423) 2 + 17 (398) 2 = ---------------------------- = ---------------------------- 10 + 17 – 2 10 + 17 – 2 = 423.42 1456 – 1280 10 x 17 176 1456 – 1280 10 x 17 176 t = ----------------- ----------- = --------- x 2.51 = 1.043 423.42 10 + 17 423.42 423.42 10 + 17 423.42

29. V = 10 + 17 – 2 = 25 V = 10 + 17 – 2 = 25 V = 25 t 0.05 = 2.06 V = 25 t 0.05 = 2.06 The calculated value / table value we accept the null hypothesis that two batches do not differ significantly in their means. The calculated value / table value we accept the null hypothesis that two batches do not differ significantly in their means.

30. The Difference Test If we are interested to find the effect of a medicine, or the effect of coaching on a particular group of students, then we apply difference test. The t value is obtained by the formula If we are interested to find the effect of a medicine, or the effect of coaching on a particular group of students, then we apply difference test. The t value is obtained by the formula  d n  d n t = --------- S V = n – 1 V = n – 1 Where d = mean of the difference = standard deviation of the differences s = standard deviation of the differences

31. ∑ ( d –  d ) 2 ∑ d 2 - n (  d ) 2  ∑ ( d –  d ) 2 ∑ d 2 - n (  d ) 2  S = ------------- or -------------------- n – 1 n - 1 n – 1 n - 1

32. A certain stimulus administered to each of 12 patients resulted in the following changes in blood pressure 5, 2, 8, -1, 3, 0, -2, 1, 5, 0, 4, 6 A certain stimulus administered to each of 12 patients resulted in the following changes in blood pressure 5, 2, 8, -1, 3, 0, -2, 1, 5, 0, 4, 6 Can it be concluded that the stimulus will in general be accompanied by an increase in blood pressure? Can it be concluded that the stimulus will in general be accompanied by an increase in blood pressure? Let us take the hypothesis that the stimulus does not increase the blood pressure Let us take the hypothesis that the stimulus does not increase the blood pressure  d n  d n t = -------- t = -------- s

33. dd2d2 525 24 864 1 39 00 -24 11 525 00 416 636 ∑d = 31 ∑d 2 = 185

34. ∑ d 31 ∑ d 31  d = --------- = ------- = 2.583 n 12 n 12 ∑d 2 - n (  d ) 2 185 -12 (2.583) ∑d 2 - n (  d ) 2 185 -12 (2.583)2 S = ------------------ = --------------------- = 3.09 n -1 11 n -1 11  d n 2.583  d n 2.583 t = ---------- = --------- 12 = 2.896 S 3.09 S 3.09 V 12 - 1 = 11 t 0.05 = 2.2 The calculated value > table value of t V 12 - 1 = 11 t 0.05 = 2.2 The calculated value > table value of t Hence Null Hypothesis is rejected. We therefore conclude that the stimulus is accompanied by an increase in blood pressure. Hence Null Hypothesis is rejected. We therefore conclude that the stimulus is accompanied by an increase in blood pressure.