f and t test

Biostatics F test and T test

1. ECONTENT ST. THOMAS COLLEGE, BHILAI PREETI VERMA ASSISTANT PROFESSOR IN BIOTECHNOLOGY ST. THOMAS COLLEGE, BHILAI F - TEST AND T - TEST

2. SYNOPSIS Introduction History Hypothesis Types of error t – Test Testing procedure Types of t – test Example of t – test t- Table f- Test one and two way ANOVA Examples of f – test f- Table Applications of test of significance Conclusions References

3. Introduction In normal English, " significant " means important , while in Statistics "significant" means probably true (not due to chance ). A procedure to assess the significance of difference between two samples drawn from the population or from closely related population is known as the test of significance. Testing of hypothesis can be defined as “ a procedure that helps in ascertaining the likelihood of hypothetical parameter of a population being correct by using sample statistics ” The tests of significance involves t – test and f – test .

4. history The T- statistics was introduced in 1908 by William Sealy Gosset . The name “ f – test ” was coined by George W. Snedecor , in honor of Sir Ronald A. Fisher . Fisher initially developed the statistic as the variance ratio in the 1920s .

5. Hypothesis The two hypotheses in a statistical test are normally referred to as : Null hypothesis : A no difference hypothesis i.e. there is no difference between sample statistic and population parameter is null hypothesis. Denoted by H 0 . H : µ = x ˉ 2. Alternative hypothesis : “Any hypothesis which is complementary to the null hypothesis is called alternative hypothesis.” Rejection of H leads to the acceptance of alternative hypothesis which is denotes by H 1 or H a .

6. Types of error There are two possible types of errors in the test of a hypothesis: Type I error : Rejection of null hypothesis which is true. = α Type II error : Acceptance of null hypothesis which is false. = β Reject H Accept H H is true Type I error Correct decision H is false Correct decision Type II error

7. t - test This test is used to test whether sample mean is equal to population mean or not in case of small samples . Mean difference Standard error (S.E.) t =

8. Testing procedure The procedure for testing of hypothesis is as follows: 1. Set up null hypothesis H . 2. Set up alternate hypothesis H 1. 3. Check whether sample size is large or small. 4. Choose appropriate formula for standard error and calculate it. 5. Calculate the test statistic t. 6. Compare the calculated value with the table value with certain level of significance.

9. Types of t - test There are mainly three types of t – test . 1. One sample t – Test : A single sample t-test is used when there is only one independent sample that is being compared to a population of interest. The formula for calculating a t-score is: H = m = m (given in object) t = mean difference / SE where SE is the standard error of the mean. The standard error is calculated: SE = S /√ n - 1 where S is the sample standard deviation and n is the sample size and df is the degrees of freedom(n-1).

10. 2. Matched pairs t – Test : A paired samples t-test is a type of t-test where a single sample of participants is used more than once on some factors at two point in time. A paired samples t-test is calculated using the formula: H = d = 0 (given in the object) d = effect of treatment = observation after treatment – observation before treatment t = d / √s 2 / n s 2 = 1 / n – 1 [ Ʃd 2 – ( Ʃd ) 2 / n] where s is sample SD, n is sample size and d.f . = (n – 1) 3. Two sample t- Test : A two samples t-test is used to determine differences between the means of two distinct samples within a population. The formula for the independent samples t-test is: H = m = m (given in object) t = mean difference / SE SE = √s 2 [1 / n 1 + 1 / n 2 ] s 2 = n 1 s 2 1 + n 2 s 2 2 / n 1 + n 2 – 2 Where d.f . = n 1 + n 2 – 2

11. Example of t - test Ten rats were fed with rice in 1 st months and body weights of the rats were recorded. In the next month, they were fed with grams and their weights were measured again. The respective weights of ten rats in two months are as follows :   Weights in 1 st month 50 60 5 8 52 51 62 58 55 50 65 Weights in 2 nd month 56 58 68 61 56 59 64 60 50 62

12. Solution – H : Weights of 1 st and 2 nd months are equal. s. no. Weights in 1 st month X 1 Weights in 2 nd month X 2 Difference X 1 – X 2 = d d 2 1. 50 56 -6 36 2. 60 58 +2 4 3. 5 8 68 -10 100 4. 52 61 -9 81 5. 51 56 -5 25 6. 62 59 +3 9 7. 58 64 -6 36 8. 55 60 -5 25 9. 50 50 10. 65 62 +3 9 Ʃd = -33 Ʃd 2 = 325

13. d = -33 / 10 = - 3.3 s 2 = 1 / 10 -1 [325 – (-33 * - 33) / 10] = 1 / 9 [325 – 1089 / 10] = 1 / 9 [325 – 108.9] = 24 | t | = 3.3 / √24 / 10 = 3.3 / √2.4 = 3.3 / 1.55 = 2.13   Table value of t at 5% level of significance for 9 d.f . is 1.833 Calculated value > table value 2.13 > 1.833 Hence reject H : Weights of 1 st and 2 nd months are not equal.   d = Ʃd / no. of observation s 2 = 1 / n – 1 [Ʃd 2 – ( Ʃd ) 2 / n] t = d / √s 2 / n

14. t - table

15. f - test f- test is the ratio of between group mean square and within group mean square. The formula for calculating a f-score is F = Mean square between samples Mean square within samples

16. One-way ANOVA In statistics, one-way analysis of variance (abbreviated one-way ANOVA) is a technique used to compare means of two or more samples (using the F distribution). This technique can be used only for numerical data. In one way classification, only one factor is influenced. Two-way ANOVA The two-way ANOVA is an extension of the one-way ANOVA. The "two-way" comes because each item is classified in two ways. Therefore, analysis of variance can be used to test the effects of two factors simultaneously. One & two way anova

17. Examples of f - test Example1: From the data given below, find out whether the mean of the three sample differ significantly or not. Sample 1 Sample 2 Sample 3 20 19 13 10 13 12 17 17 10 17 12 15 16 9 5

18. Sample 1 Sample 2 Sample 3 20 19 13 10 13 12 17 17 10 17 12 15 16 9 5 Ʃx = 80 70 55 Null hypothesis: H : There is no significance difference in the means of the three samples. Solution:

19. Ʃx = Ʃx c1 + Ʃx c2 + Ʃx c3 = 80 + 70 + 55 = 205 Ʃx 2 = Ʃx 2 c1 + Ʃx 2 c2 + Ʃx 2 c3 = 1334 + 1044 + 663 = 3041 (A) ( Ʃx ) 2 / n c = Ʃx 2 c1 / n c1 + Ʃx 2 c2 / n c2 + Ʃx 2 c3 / n c3 = 80 2 / 5 + 70 2 / 5 + 55 2 / 5 = 1280 + 980 + 605 = 2865 (B) C.F . = ( Ʃx ) 2 / n = (205) 2 / 15 = 42025 / 15 + 2801.67 (D)

20. Analysis of the variance table Source of variation d.f. SS MS = SS / df F Between samples c - 1 3-1 = 2 (B – D) 2865 – 2801.67 = 63.33 (B – D) / c – 1 31.67 / 2 = 31.67 31.67 / 14.67 = 2.16 Within samples c (r - 1) 3(5-1) = 12 (A – B) 3041 – 2865 = 176 (A – B) / c(r -1) 176 / 12 = 14.67 total cr – 1 15 – 1 = 14 (A – D) 3041 – 2801.67 = 239.33 Conclusion : calculated F value < tabulated F value 2.16 < 3.9 ( 5%) The null hypothesis is accepted. The mean of various samples do not differ significantly among themselves.

21. Example2: In an experiment, the mean yields of three rice varieties grown with four nitrogen rates were recorded. Analyze the data using the test of analysis of variance to determine whether there is any difference in the mean yield of three varieties with four nitrogen doses. The results are given in the following table . Nitrogen rate kg/ha V1 V2 V3 4.50 5.01 6.11 30 4.30 6.17 6.92 60 5.60 6.37 6.37 90 5.21 6.48 6.48

22. Solution: Null hypothesis: H : Let us assume that there is no difference in the mean yield of the three varieties . Nitrogen rate kg/ha V1 V2 V3 Total 4.50 5.01 6.11 15.62 30 4.30 6.17 6.92 17.39 60 5.60 6.37 6.37 19.24 90 5.21 6.48 6.48 19.91 Ʃx 19.61 24.39 28.16 72.16

23. Ʃx 2 = 97.24 + 150.23 + 199.85 = 447.62 (A) ( Ʃx ) 2 / n c = 19.61 2 / 4 + 24.39 2 / 4 + 28.16 2 / 4 = 443.11 (B) ( Ʃx ) 2 / n r = 15.62 2 / 3 + 17.39 2 / 3 + 19.24 2 / 3 + 19.91 2 / 3 = 437.66 (C) C.F. = ( Ʃx ) 2 / n = 72.16 2 / 12 = 5207.70 / 12 = 433.92 (D)

24. Analysis of variance table Source of variation d.f. SS MS = SS / df F Between varieties ( columns) c - 1 4-1 = 3 (B – D) 443.11 – 433.92 = 9.19 (B – D) / c – 1 9.19 / 3 = 3.06 3.06 / 0.13 = 25.34 Between treatments (rows) (r - 1) 3 -1 = 2 (C – D) 437.66 – 433.92 = 3.74 (C – D) / (r -1) 3.74 / 2 = 1.87 1.87 / 0.13 = 14.38 Residual (c -1) (r – 1) (4 -1) (3 – 1) = 6 (A – D) – [(B – D) + (C – D)] 13.7 – (9.19 + 3.74) = 0.77 0.77 / 6 = 0.13 Total cr – 1 4*3 – 1 = 11 (A – D) 447.62 – 433.92 = 13.7

25. Conclusion: ( i ) F for between treatments: Tabulated F value = 9.8, p = 0.01 df = n 1 = 3; n 2 = 6 calculated F value > tabulated F value 25.34 > 9.8 The null hypothesis, stating that the three varieties have the identical mean yield is rejected. (ii) F for between nitrogen rate: Tabulated F value = 10.9, p = 0.01 df = n 1 = 2; n 2 = 6 The calculated f (14.38) value for, between the fertilizer doses exceeds the tabulated value. Thus, the variation in four nitrogen doses had significant effect on the mean yield and the null hypothesis is rejected at 1% level.

26. applications t– test is used To study the role of a factor or cause when the observations are made before and after its play. To compare the effect of two drugs, given to the same person. To compare the results of two different laboratory technique. f – test is used To test the two independent samples (X and Y) have been drawn from the normal population with same variance. Whether the two independent estimates of the population variance are homogenous or not.