Sample B_Book Typesetting

1,221 views
1,175 views

Published on

Book Typesetting

Published in: Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
1,221
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
6
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Sample B_Book Typesetting

  1. 1. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 1 of 17 May 31, 2009 17:29 13 SIGNIFICANT TESTS A fter any experiment we get some results, but we are not sure about this result whether the result occurred by chance or a real difference. That time to find truth we will use some statistical tests, these tests are termed as, ‘Tests of Significance’. Selection of Statistical Tests: The selection of the appropriate statistical test is depends upon: 1) The scale of measurement e.g. Ratio, Interval. 2) The number of groups e.g. One, Two or More. 3) Sample size e.g. If the sample size is less than 30. Students ‘t’ test is to be used. 4) Measurements e.g. Repeated or Independent measurements. Selection of Test of Significance: Scale Two groups Three/More groups Independent Repeated Independent Repeated Interval Z test Z test ANOVA test ANOVA and Ratio t test t test (F test) (F test) Ordinal Median test Wilcox an Median Friedman Mann test Kruscal test test Whitney Nominal X2 test Me Nemar Chi–Square Cochron’s test Test test For application of test sample should be selected randomly. 1
  2. 2. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 2 of 17 May 31, 2009 17:29 2 SIGNIFICANT TESTS There are two types of tests: 1) Parametric Tests 2) Non – Parametric Tests 1. Parametric Test: When quantitative data like Weight, Length, Height, and Percentage is given it is used. These tests were based on the assumption that samples were drawn from the normally distributed populations. E.g. Students t test, Z test etc. 2. Non – Parametric Test: When qualitative data like Health, Cure rate, Intelligence, Color is given it is used. Here observations are classified into a particular category or groups. E.g. Chi square (x2 ) test, Median tests etc. I) T – Test: W.S. Gosset investigated this test in 1908. It is called Student t – Test because the pen name of Dr. Gosset was student, hence this test is known as student’s t – test. It is also called as ‘t- ratio’ because it is a ratio of difference between two means. Aylmer Fisher (1890–1962) developed students ‘t’ test where sam- ples are drawn from normal population and are randomly selected. After comparing the calculated value of ‘t’ with the value given in the ‘t’ table considering degree of freedom we can ascertain its significance. Patients Before After treatment (B) treatment (A) 1 2.4 2.2 2 2.8 2.6 3 3.2 3.0 4 6.4 4.2 5 4.3 2.2 6 2.2 2.0 7 6.2 4.8 8 4.2 2.4 Is the testing reliable?
  3. 3. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 3 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 3 Solution: Patients Before After Difference D2 treatment (B) treatment (A) B−A=D 1 2.4 2.2 0.2 0.04 2 2.8 2.6 0.2 0.04 3 3.2 3.0 0.2 0.04 4 6.4 4.2 2.2 4.84 5 4.3 2.2 2.1 4.41 6 2.2 2.0 0.2 0.04 7 6.2 4.8 1.4 1.96 8 4.2 2.4 1.8 3.24 D = 8.3 D2 = 14.61 Here, D = 8.3 N=8 D2 = 14.61 8.3 D= = 1.0375 8 ∴ Standard deviation of the different between means. ( D)2 D2 − n = N−1 (8.3)2 14.61 − 8 = 7 14.61 − 68.89 8 = 7 14.61 − 8.6112 ∴ S.D. = 7 √ = 0.8569 ∴ S.D. = 0.9257
  4. 4. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 4 of 17 May 31, 2009 17:29 4 SIGNIFICANT TESTS Now, standard error of the difference (SED ) SD 0.9257 0.9257 .= √ = √ = N 8 2.8284 ∴ S.E. = 0.3272 D 1.0375 ∴t= = = 3.1708 SED 0.3272 Here, the calculated value for ‘t’ exceeds the tabulated ‘t’ value at p = 0.05 level with 7df. Therefore the glucose concentration by the patients after treatment is not significant. Utility: It is widely used in the field of Medical science, Agriculture and Veterinary as follows: r To compare the results of two drugs which is given to same individuals in the sample at two different situations? E.g. Effect of Bryonia and Lycopodium on general symptoms like sleep, appetite etc. r It is used to study of drug specificity on a particular organ / tissue / cell level. E.g. Effect of Belberis Vulg. on renal system. r It is used to compare results of two different methods. E.g. Estimation of Hb% by Sahlis method and Tallquist method. r To compare observations made at two different sites of the same body. E.g. compare blood pressure of arm and thigh. r To study the accuracy of two different instruments like Ther- mometer, B.P apparatus etc. r To accept the Null Hypothesis that is no difference between the two means. r To reject the hypothesis that is the difference between the means of the two samples is statistically significant. F – Test: A statistician R.A. Fisher introduces it. That is why it is also called as Fisher’s test (F – test) test.
  5. 5. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 5 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 5 Definition: It is the ratio of two independent chi-square variables which is derived by dividing each by its corresponding degree of freedom. Ψ1 2 V1 ∴F= Ψ2 2 V2 Here, two variances are derived from two samples. The values in each group are to be normally distributed. Therefore the variation of each value around its group mean that is error is remain independent of each value provided the variances within each group should be equal for all groups. Calculation for F Test: Tests of hypothesis about the variance of two populations:- Steps:- 1) Null hypothesis should be H0 = 61 2 = 62 2 and Alternative hypo- thesis H0 = 61 2 = 62 2 (two tailed test) 2) Calculation of F test statistic:- 61 2 F= if, 61 2 ≥ 62 2 62 2 OR 62 2 F= if, 62 2 ≥ 61 2 61 2 3) Take the level of significance α = 0.05 If α is not known. Rules for Significance: 1) If calculated F < tabled Fα2 then accept H0 . 2) If, calculated F > tabled Fα2 then reject H0 .
  6. 6. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 6 of 17 May 31, 2009 17:29 6 SIGNIFICANT TESTS Steps for One Tailed Test: 1. Set the Null Hypothesis (H0 ): 61 2 > 62 2 in such a way that the rejection appeases in the upper tail by numbering the population variance. Format: H0 : 62 2 < 61 2 If, H0 is of the form 61 2 < 62 2 then we calculate F = 61 2 /62 2 which has F – distribution but with n2 − 1 d.f. in the numerator and n − 1 d.f. in the denominator. 2. Calculation of test statistics: 61 2 ∴ F statistics = 62 2 3. Set the level of significance α= 0.05 if value of α is not known to us. 4. Rules for Significance: If calculated F α table F2 α then accept the null hypothesis H0 and reject H0 if calculated F > table Fα. Example: 1) Random samples are drawn from the two sets of students and the following results were obtained. Sample A: 10, 12, 18, 14, 16, 20. Sample B: 14, 16, 20, 18, 21, 22. Find the variance of two populations and test whether the two samples have same variance. Solution: Null hypothesis H0 = 6A 2 = 6B 2 that is the two samples have the same variance. Alternative hypothesis H0: 6A 2 = 6B 2 (Two tailed test)
  7. 7. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 7 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 7 Calculation of test statistic: We have the following table to evaluation 6A 2 and 6B 2 A A – A (A – A)2 B – B (B – B)2 (A – 15) (A – 15)2 B (B – 18.5) (B – 18.5)2 10 −5 25 14 −4.5 20.25 12 −3 9 16 −2.5 6.25 18 3 9 20 1.5 2.25 14 −1 1 18 −0.5 0.25 16 1 1 21 2.5 6.25 20 5 25 22 3.5 12.25 A = 90 (A − A)2 B = 111 (B − B)2 = 70 = 47.5 90 We know, A= = 15. 6 Rule of Significance: 1) If, the calculated value of ‘t’ is higher than the value given at P = 0.05 (5% level) in the table it is significant. 2) If the calculated value of ‘t’ is less than the value given in ‘t’ table it is not significant. Degree of Freedom: It is the quantity in a series which is one less than the independent number of observations in a sample is called – Degree of freedom. E.g. In unpaired t test df = N − 1 and in paired t test df = N1 + N2 − 2 (Where, N1 and N2 are the number of observations.) There are two types of t – test: a) Unpaired t – Test. b) Paired t – Test. A] Unpaired t – Test: Indications for Unpaired t – Test: i) When, samples drawn from two population OR
  8. 8. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 8 of 17 May 31, 2009 17:29 8 SIGNIFICANT TESTS When, unpaired data of independent observations of two different groups are given. Calculation for Significance Tests: Following steps are taken to test the significance of difference. Steps: 1) Find the observed difference between means of two samples. (X1 − X2 ) 2) Calculate the SE of difference between means or SEn . 61 2 62 2 That is S (X1 − X2 ) = + N1 N2 3) Apply formula for t (X1 − X2 ) That is t = 61 2 62 2 + N1 N2 4) Find the degree of freedom. Example: 1) Raulfia Ø is given to each of the 8 patients resulted in the following changes in the Blood pressure from normal. −5, 3, 4, −2, 7, 3, 0, 2 Calculate by students ‘t’ – test whether changes is significant or not. Solution: Calculation of mean: X ∴X= N 12 = 8 = 1.5
  9. 9. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 9 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 9 Calculation of S.D: X X–X=x x2 −5 −5 − 1.5 = −6.5 42.25 3 3 − 1.5 = 1.5 2.25 4 4 − 1.5 = 2.5 6.25 −2 −2 − 1.5 = −3.5 12.25 7 7 − 1.5 = 5.5 30.25 3 3 − 1.5 = 1.5 2.25 0 0 − 1.5 = −1.5 2.25 2 2 − 1.5 = 0.5 0.25 N=8 x2 = 98 x2 ∴ = S.D. = N−1 98 √ = = 14 = 3.74 7 √ X× N Now, t = S.D. 1.5 × 2.82 = 3.74 = 1.13 Degree of freedom = N − 1 = 8−1 =7 Here, calculated value of ‘t’ is less than the given value in t table hence the difference between the two means is significant. E.g. 2) The weight of an untreated group of six persons are 60kg, 40kg, 45kg, 50kg, 65kg, 70kg. The weight of another group persons from the same population other treatment with Phytolacca drug was obtained as, 45kg, 35kg, 40kg, 45kg, 60kg, 65kg and 45 kg. Apply t test to find out significance of difference between means of two groups.
  10. 10. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 10 of 17 May 31, 2009 17:29 10 SIGNIFICANT TESTS Solution: Calculation of Mean: Untreated Persons Weight X1 X1 − X1 = x x1 2 1 60 60 − 55 = 5 25 2 40 40 − 55 = −15 225 3 45 45 − 55 = −10 100 4 50 50 − 55 = −5 25 5 65 65 − 55 = 10 100 6 70 70 − 55 = 15 225 X1 = 330 x1 2 = 700 X1 ∴ Mean of first group = X1 = N1 330 = 6 = 55 Treated persons Weight X2 X2 − X2 = x2 x2 2 1 45 45 − 47.85 = −2.85 8.12 2 35 35 − 47.85 = −12.85 165.12 3 40 40 − 47.85 = −7.85 61.62 4 45 45 − 47.85 = −2.85 8.12 5 60 60 − 47.85 = 12.15 147.62 6 65 65 − 47.85 = 17.15 294.12 7 45 45 − 47.85 = −2.85 8.12 N2 = 7 X2 = 335 x2 2 = 692.84 X2 And Mean of 2nd group = X2 = N1 335 = 7 = 47.85
  11. 11. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 11 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 11 Now, calculate combined S.D. by applying formula: (x1− X1 )2 + (x2 − X2 )2 ∴ Combined S.D. = N1 + N2 − 2 √ 700 + 692.84 = 6+7−2 1392.87 = 11 ∴ S.D. = 11.25 Calculation of ‘t’ by applying following formula: X1 − X2 ∴t= N1 + N2 6 N1 + N2 55 − 47.85 = 6+7 11.25 6+7 7.15 = 13 11.25 × 13 7.15 ∴t= √ 11.25 × 1 7.15 = 11.25 × 1 ∴ t = 0.635 Here, the calculated value for t. (0.635) is more than that given in the ‘t’ table for degree of freedom N = (N1 + N2 − 2 = 6 + 7 − 2 = 11). Hence, the difference between two means is not significant.
  12. 12. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 12 of 17 May 31, 2009 17:29 12 SIGNIFICANT TESTS b] Paired ‘t’– Test: Indications for Paired t – Test: i) When paired data of independent observation from one sample only given. Calculation for Significance Test: Following steps are to be taken to test the significance of difference. Steps: 1) Find the difference in each set of paired observations before and after treatment (X1 − X2 ) = x. 2) Calculation of Mean of Difference that is x. S.D. 3) Calculate S.D. of difference and S.E. of Mean from the same √ N X−0 X 4) Apply formula for ‘t’ that is t = = 5 SEd √ N 5) Find the degree of freedom. Example: 1. Two research centers carry out independent estimates of calcium carbonate content for water made by a certain firm. A sample is taken from each place and sent to the two centers separately. They obtain the following results. Percentage of Calc. Carbonate content in water. Place No. 1 2 3 4 Center A 8 5 6 3 Center B 6 6 5 4 Is the testing reliable? Solution: Ho; µD = 0 Here, the testing will be reliable if the mean difference between the results from the two-research centers does not differ significantly form zero. So we assume the hypothesis that the observed differences are the random observations from a population with mean zero.
  13. 13. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 13 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 13 Calculation of mean difference and standard error: Place No. Center Diff. of results Center A Center B D=B−A D2 1 8 6 −2 4 2 5 6 1 1 3 6 5 −1 1 4 3 4 1 1 Total 22 21 −1 7 −1 Here, D = 4 = −0.25 and µ = 0 ( D)2 (D − D)2 = D2 − N (−1)2 = 7− 4 (−2) = 7− 4 = 7 − (−0.5) = 6, 5 (D − D) ∴ S.E of D = N(N − 1) 6.5 = 4(4 − 1) 6.5 = 12 = 0.735 D−µ −0.25 − 0 ∴t= = = −0.340 S.E. of D 0.735 D.f. = 4 − 1 − 3.
  14. 14. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 14 of 17 May 31, 2009 17:29 14 SIGNIFICANT TESTS Since the observed value of the t = (− 0.340) is less then the value of t at 5% level of significance for 3 df. So it is non significant. Hence the hypothesis will be accepted that is the testing is reliable. Example 2. The effect of Synz. Jamb. drug on 8 patients showed concentration of glucose (mg/hr) after 24 hrs as follows: 111 x B= = 18.5 we know x = n 6 (A − A)2 70 ∴ 61 2 = = = 14 n1 − 1 5 (B − B)2 47.5 ∴ 62 2 = = = 9.5 n2 − 1 5 62 2 9.5 ∴ Test statistics: F = = = 0.6785 61 2 14 Conclusion: The calculation value of F = 0.6785, < Table value F 0.05 the null hypothesis H0 is accepted. ∴ The two samples have the same variance. Testing the Homogeneity of Variance between Groups: We have following formula, Largest Variance ∴F= Smallest Variance Where the ratio is directly proportional to its significance. E.g. Standard Deviation was in three groups of students for their marks in statistics conclude the variance between groups significantly differing from each other. Sr. No. Group (1) Group (2) Group (3) 1 Sample size N=10 N= 13 N = 15 2 Standard deviation 0.2757 1.213 0.5512 3 Variance (SD)2 0.076 1.47 0.303
  15. 15. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 15 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 15 Here, the largest variance is group (2) and smallest variance is group (1). Largest variance ∴ Variance Ratio (F) = Smallest variance 1.47 = 0.076 df = 13 − 1 = 12 And = 10 − 1 = 9 Find out at df 12, 9 table F value, If the calculated F is greater than tabulated value it indicates that variance are not homogenous and vice versa. II] Non Parametric Tests: 1. Chi–Square Test (x2 ): It is one of the non – parametric tests where qualitative data is considered. It is used to test the significance of overall deviation between the Observed and Expected frequencies. Chi-square is derived from the Greek letter – Chi-x. Prof. A.R. Fisher developed this test in 1870. It was Karl Pearson who improved Fisher’s Chi-square test in its latest form in 1900. It is the test of significance of overall deviation square in the observed and expected frequencies divided by expected frequencies. (0 − E)2 x2 = E OR (fo − fe)2 x2 = fe Where, O or fo = Observed frequency. E or fe = Expected frequency. N = Total number of observations. Rules for Significance: 1) If the tabular value is lower than the calculated value then the results are significant.
  16. 16. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 16 of 17 May 31, 2009 17:29 16 SIGNIFICANT TESTS 2) If fo = fe then the value of x2 will be zero (but due to chance error this never happens). Example: 1) There are two factors showing dominance X and Y. Suppose in A the progeny were in the ratio of XY = 436, Xy = 122, xY = 120 and xy = 64 out of 646 individuals. Test the hypothesis that an ‘A’ gives 6: 2: 2: 1 isolation. Solution Steps: 1) XY = observed = 436. 6 Expected = 646 × = 352.36 11 O − E = 436 − 352.36 = 83.64 (O − E)2 = (83.64)2 = 6995.6496 (O − E)2 6995.6496 = = 19.85 E 352.36 2) Xy = observed = 122, 2 Expected = 646 × = 117.45 11 O − E = 122 − 117.45 = 4.55 (O − E)2 = 20.7025 (O − E)2 20.7025 = = 0.1762 E 117.45 3) xY= observed = 120. 2 Expected = 646 × = 117.45 11 O − E = 120 − 117.45 = 2.55 (O − E)2 = 6.5025 (O − E)2 6.5025 = = 0.05536 E 117.45
  17. 17. ch-13 NKR/LaTeX Sample (Typeset by TYINDEX, Delhi) 17 of 17 May 31, 2009 17:29 SIGNIFICANT TESTS 17 4) xy = observed = 64. 1 Expected = 646 × = 58.7272 11 O − E = 64 − 58.7272 = 5.2728 (O − E) = 5.2728 (O − E)2 27.8024 = = 0.4734 E 58.7272 Now, using the formula, (O − E)2 X2 = E = 19.85 + 0.1762 + 0.05536 + 0.4734 = 20.554 In all such cases degree of freedom (df) will be n = k – 1 (where k is the number of classes). Thus we have in this case degree of freedom ‘n’ = 4 – 1 = 3 Now, table value of x2 is 7.815 t 0.05 for 3 degree of freedom, which is much less than the obtained value that is 20.554 So we reject the hypothesis of 6 : 2 : 2 : 1 in this case Utility: r It is used for comparisons with expectations of the Normal, Binomial and Poisson distributions and Comparison of a sample variance with population variance. r It is used for testing the Homogeneity, Correlation and Pro- portion and Independence of sample variances, Attributes and Expectation of ratio. r It is useful in the field of Genetics for detection of linkage.

×