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Comparing Two Samples: Part I

Comparing Two Samples: Part I. Measurements (data). Descriptive statistics. Data transformation. Normality Check Frequency histogram (Skewness & Kurtosis) Probability plot, K-S test. YES. NO. Mean, SD, SEM, 95% confidence interval. Median, range, Q1 and Q3 . Data transformation.

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Comparing Two Samples: Part I

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  1. Comparing Two Samples: Part I

  2. Measurements (data) Descriptive statistics Data transformation Normality Check Frequency histogram (Skewness & Kurtosis) Probability plot, K-S test YES NO Mean, SD, SEM, 95% confidence interval Median, range, Q1 and Q3 Data transformation Check the Homogeneity of Variance Non-Parametric Test(s) For 2 samples: Mann-Whitney For 2-paired samples: Wilcoxon For >2 samples: Kruskal-Wallis Sheirer-Ray-Hare NO YES Parametric Tests Student’s t tests for 2 samples; ANOVA for  2 samples; post hoc tests for multiple comparison of means

  3. Procedures for comparing two samples • Test normality of the data; if passed, then • Compare the measurements both of location (central value) and dispersion (spread or range) of the thickness indices between the two sites • If there is a difference only in the measure of location, a parametric statistical test based upon the difference between the two sample means

  4. SX = Standard error of mean = SD/n One sample t test: t = (X - )/SX • The 2-tailed t test for significant difference between a mean longevity of horses and a hypothesized population mean of  = 22 yr • Ho:  = 22 year • Age at death (in year) of 25 horses: t = (24.23 - 22)/ (4.25/25) = 2.624  = n - 1 = 25 - 1 = 24 t 0.05 (2), 24 = 2.064 Thus reject Ho 0.01<p<0.02

  5. Confidence interval for mean • A researcher often needs to know how close a sample mean (x) is to the population mean (). For example, the amount of dissolved/dispersed petroleum hydrocarbons (DDPH) in the upper ocean. It is impossible to sample the entire ocean so the population mean at any time is unknown. If data of the population follow the normality, the sample mean will be normally distributed. • The standard deviation (or standard error ) of the mean can be obtained by: SX = S.E.M. = S/n • In general, the population mean lies within one standard deviation (S) of the sample mean with about 68% confidence. It is usual to quote 95% confidence intervals by multiplying S.E.M. with the (1) appropriate value of z, namely 1.96 (for n >30) or (2) appropriate value of t, df = n -1 (for n < 30).

  6. Confidence interval for mean • e.g. 50 measurements of DDPH in the upper ocean were made. The sample mean = 4.75 ppb and S = 3.99 ppb Then SX = 3.99/ 50 = 0.5643 The 95% confidence intervals are obtained as X  z 0.05SX = 4.75  (1.96)(0.5643) = 4.75  1.105 ppb i.e. 5.855 and 3.633 ppb • e.g. Only 10 measurements of DDPH in the upper ocean were made. The sample mean = 78.2 ppb and S = 38.6 ppb Then SX = 38.6/ 10 = 12.206 The 95% confidence intervals are obtained as X  t 0.05, n-1SX = 78.2  (2.262)(12.206) = 78.2  27.61 ppb

  7. Two-tailed and one-tailed tests • If we do not know which of the means is greater than the other, we reject extreme values in either direction (i.e. HA: a  b); this procedure is known as a two-tailed test. • However, if sufficient information is available for us to specify the direction of the population means, we could frame the alternative hypothesis as HA: a > b or a < b • For example, comparison of total organic carbon (TOC) concentrations in the sewage effluent before and after being upgraded to an advance treatment. In this case, a reduced in TOC would be expected after improvement of the sewage treatment, and therefore, HA: before > after • The corresponding critical value of z for 1-tailed tests is slightly different from those for 2-tailed tests (see Table B3).

  8. The difference between two sample means with limited data • If n<30, the above method gives an unreliable estimate of z • This problem was solved by ‘Student’ who introduced the t-test early in the 20th century • Similar to z-test, but instead of referring to z, a value of t is required (Table B3) • df = 2n - 2 for n1 = n2 • For all degrees of freedom below infinity, the curve appears leptokurtic compared with the normal distribution, and this property becomes extreme at small degrees of freedom.

  9. Comparison of two samples Mean measured unit A B

  10. Two-sample t test Measured unit A B Error bar = 2 SD

  11. Two-sample t test Measured unit A B Error bar = 2 SD

  12. Measured unit A B Error bar = 2 SD

  13. Importance of Equal Variance • If the measure of dispersion (i.e. homogeneity of variance) also differs significantly between the samples: • data transformation procedure: if there was no significant difference between the variances of transformed data, then a parametric test could be applied; otherwise you should consider • non-parametric test or Welch’s approximate t’ (Zars p. 128-129)

  14. Assumptions for z and t tests • For the t and z tests to be valid: • the measurements should be at least approximately normally distributed and • with similar sample variances (i.e., homoscedastic or homogeneity of variance) • Need to check Normality of both datasets • Homoscedasticity can be checked by a F-test The largest sample variance F ratio = The smallest variance

  15. Check for homogeneity of variance Ho: equal variance between the two samples HA: unequal variances • Given that Sa2 = 0.2898 (n = 15) and Sb2 = 0.1701 (n =12), • Then,F ratio = 0.28929/0.1701 = 1.703 • Use the F table (Table B4, Zar 99, App.34), Critical F 0.05(2), 14, 11 = 3.36 > 1.703 Accept Ho • If Ho rejected, then transform data and redo F test • If still failed, non-parametric or Welch’s approximate t’

  16. An example • Birds’ eggshells are thought to be influenced by acid rain which reduces the egg thickness index (egg shell mass/ surface area; mg cm-2) • We investigate gulls’ eggs at two nesting sites: a control site and a site affected by acid rain • Taking a single egg at random from a number of nests at each site; determining the egg thickness index

  17. Two sample Z test • Ho: no significant difference in the the thickness indices between the two sites • i.e. the thickness indices belong to the same population (or probability density function). Then the population means for the two sites will be the same and their difference will be zero. • Based on the central limit theorem, a collection of sample means will follow a normal distribution. It can also be shown that the distribution of the difference of two means (Xa - Xb) will also be normally distributed. • Recall the distribution of z: if (Xa - Xb) can be divided by an appropriate standard deviation, a value of z can be calculated:

  18. Two sample Z test • The standard deviation of (Xa - Xb) = [(1/n)(Sa2 + Sb2)] • Then, Z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)] • This equation can be used provided the number of measurements in each sample is reasonably large (n > 30), the measurements are approximately normally distributed and n in each sample is the same. • Ho: a = b ; HA: a  b • If the sample means are similar, the z value will be small and Ho will be accepted. The critical values of z can be obtained from the t table, Table B3, (df = ).

  19. Two sample Z test • The following data were obtained at two gull nesting sites. 50 eggs were taken at random from each site, with 1 egg taken randomly from each nest. Egg thickness index results (mg cm-2) Site a (control) b n 50 50 mean 33.2 31.89 S2 16.41 17.39 Ho: a = b ; HA: a  b z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)] = (33.2-31.89)/[(1/50)(16.41 + 17.39)] z = 1.593 As critical z=0.05, df =  = 1.96, accept Ho. Remember to always check the homogeneity of variance before running the t test.

  20. Exercise • The following data were obtained at two gull nesting sites. 50 eggs were taken at random from each site, with 1 egg taken randomly from each nest. Egg thickness index results (mg cm-2) Site a (control) b n 50 50 mean 33.2 27.8 S2 16.41 15.21 Test the Ho: a = b (HA: a  b) using the two sample z test z = (Xa - Xb)/[(1/n)(Sa2 + Sb2)] Please do it later ! Remember to always check the homogeneity of variance before running the t test.

  21. Example A t test with equal measurements in each sample (na = nb) SS = sum of square = S2 ×υ sp2 = (SS1+ SS2) / (υ1+ υ2)= [(0.257 × 11) + (0.366 × 11)]/(11+11) = 0.312 sX1 – X2 = √(sp2/n1 + sp2/n2) = √(0.312/12) × 2 = 0.228 t = (X1 – X2) / sX1 – X2=(3.701 – 3.406) / 0.228 = 1.294 df = 2n - 2 = 22 t = 0.05, df = 22, 2-tailed = 2.074 > t observed = 1.294, p > 0.05 The calculated t-value is less than the critical t value. Thus, accept Ho. • e.g. The chemical oxygen demand (COD) is measured at two industrial effluent outfalls, a and b, as part of consent procedure. Test the null hypothesis: Ho: a = b while HA: a  b (Given that the data are normally distributed) n 12 12 mean 3.701 3.406 S2 0.257 0.366 Remember to always check the homogeneity of variance before running the t test.

  22. Example • Growth of 30-weeks old non-transgenic and transgenic tilapia was determined by measuring the body mass (wet weight). Since transgenic fish cloned with growth hormone (GH) related gene OPAFPcsGH are known to grow faster in other fish species (Rahman et al. 2001), it is hypothesized that HA: transgenic > non-transgenicwhile the null hypothesis is given as Ho: transgenic  non-transgenic

  23. Example sp2 = (SS1+ SS2) / (υ1+ υ2)= 5913.4 sX1 – X2 = √(sp2/n1 + sp2/n2) = 38.45 t = (X1 – X2) / sX1 – X2=8.29 df = 2n - 2 = 14 t = 0.05, df = 14, 1-tailed = 1.761 << 8.29; p < 0.001 The t-value is greater than the critical t value. Thus, reject Ho. Ho: transgenic  non-transgenic HA: transgenic > non-transgenic Given that mass (g) of tilapia are normally distributed. n 8 8 mean 625.0 306.25 S2 6028.6 5798.2 Remember to always check the homogeneity of variance before running the t test.

  24. Example sp2 = (SS1+ SS2) / (υ1+ υ2)= 0.519 sX1 – X2 = √(sp2/n1 + sp2/n2) = 0.401 t = (X1 – X2) / sX1 – X2=-2.470 t = 0.05, df = 6 + 7 -2 = 11, 2-tailed = 2.201 < 2.470; 0.02<p<0.05 Thus, reject Ho but accept HA. • e.g. The data are human blood-clotting time (in minutes) of individuals given one of two different drugs. It is hypothesized that Ho: a = b while HA: a  b (Given that the data are normally distributed) n 6 7 mean 8.75 9.74 S2 0.339 0.669

  25. Power of the two-sample t test • Power of two-sample t test is greatest when the number of measurements in each sample is the same (n1 = n2) • Power of two-sample t test for different numbers of measurements in each sample (n1  n2) is smaller • Whenn1  n2, effective n = 2n1n2 /(n1+n2) • e.g. n1=6, n2=7, effective n = 2(6 × 7)/ (6+7) = 6.46, which is smaller than the average of 6 and 7 (6.5) • Therefore, we should always use ‘balanced design’ where possible.

  26. Measurements (data) For comparison of two independent samples Descriptive statistics Data transformation Normality Check Frequency histogram (Skewness & Kurtosis) Probability plot, K-S test NO YES Mann-Whitney test Mean, SD, SEM, 95% confidence interval Median, range, Q1 and Q3 Data transformation F-test Check the Homogeneity of Variance Non-Parametric Test(s) For 2 samples: Mann-Whitney For 2-paired samples: Wilcoxon For >2 samples: Kruskal-Wallis Sheirer-Ray-Hare NO YES z test t tests Parametric Tests Student’s t tests for 2 samples; ANOVA for  2 samples; post hoc tests for multiple comparison of means

  27. Non-parametric methods for comparison of two independent samples Why do we need to transform the data into normal distribution for parametric tests when we can apply non-parametric tests? Parametric tests are more powerful and more sensitive than non-parametric ones.

  28. Non-parametric tests for two samples - the Mann-Whitney test • Non-parametric tests are also called ‘distribution-free’ test because they are independent of the underlying population distribution, the only assumption being independence of observations and continuity of the variable which is being measured. Most of these tests involve a ranking procedure • Mann-Whitney test is useful for two independent samples and can be used as an alternative to the t-test, particularly where the assumptions for t-test cannot be demonstrated • It can also be applied to ordinal data • It is usually assumed that the distribution of the measurements in the two samples are of the same general form (shown by frequency histogram or stem-and-leaf plot) • This test can be undertaken with any sample size. But the method of calculating the test statistic (U) depends upon the size of n.

  29. Mann-Whitney test - basic principle Case 1 35 24 23 21 19 17 15 9 4 3 1 Case 2 35 24 23 21 19 17 15 9 4 3 1 Case 3 35 24 23 21 19 17 15 9 4 3 1 Case 4 35 24 23 21 19 17 15 9 4 3 1 Unequal medians Group A Group B

  30. Example Non-parametric tests for two samples - the Mann-Whitney test • In Mann-Whitney test, the null hypothesis cannot be stated in terms of population parameters, but is defined as the equality of the medians of the populations from which the two samples are drawn. • Example: Mann-Whiney test where n is small Sample A Sample B 9 10 12 15 6 11 7 13 18 • These measurements are combined and ranked in descending order: B B B A B B A A A 18 15 13 12 11 10 9 7 6

  31. Example Example I: Mann-Whiney test (2-tailed test) • These measurements are combined and ranked in descending order : B B B A B B A A A 18 15 13 12 11 10 9 7 6 • Ho: equal medians • Summation of ranks for A score  R2 = 4 + 7 + 8 + 9 = 28 • Summation of ranks for B score  R1 = 1 + 2 + 3 + 5 + 6 = 17 • U = n1n2 + [n1(n1 + 1)/2] - R1 • where the greatest measurement (smaller sum of ranks) in either of the two groups is given as rank 1 (R1); Thus group B is R1 in this case. • U = (5)(4) + [(5)(5 +1)/2] - 17 = 18 • U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 > 18 ; thus accept Ho; i.e. equal medians

  32. Example Sample A Sample B 9 10 12 15 6 11 7 13 18 Sample A Sample B 12 18 9 15 7 13 6 11 10 Sample A Rank Sample B Rank 12 4 18 1 9 7 15 2 7 8 13 3 6 9 11 5 10 6 18 15 131211 109 7 6 Ho: equal medians U = n1n2 + [n1(n1 + 1)/2] -  R1 where the greatest measurement in either of the two groups is given as rank 1 (R1); Thus group B is R1 in this case. U = (5)(4) + [(5)(5 +1)/2] - 17 = 18 U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 > 18 ; thus accept Ho; i.e. equal medians R2 = 28 R1 = 17

  33. Example Sample A Sample B 9 10 11 15 6 12 7 13 18 Sample A Sample B 11 18 9 15 7 13 6 12 10 Sample A Rank Sample B Rank 11 5 18 1 9 7 15 2 7 8 13 3 6 9 12 4 10 6 18 15 131211 109 7 6 R2 = 29 R1 = 16 U = n1n2 + [n1(n1 + 1)/2] -  R1 U = (5)(4) + [(5)(5 +1)/2] - 16 = 19 U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19; thus reject Ho at p = 0.05; i.e. unequal medians

  34. Example Sample A Sample B 9 11 10 15 6 12 7 13 18 Sample A Sample B 10 18 9 15 7 13 6 12 11 Sample A Rank Sample B Rank 11 6 18 1 9 7 15 2 7 8 13 3 6 9 12 4 11 5 18 15 131211 109 7 6 R2 = 32 R1 = 15 U = n1n2 + [n1(n1 + 1)/2] -  R1 U = (5)(4) + [(5)(5 +1)/2] - 15 = 20 U0.05(2), 5, 4 = U0.05(2), 4, 5 = 19 < 20; thus reject Ho at p = 0.02; i.e. unequal medians

  35. Example • A variance ratio test will show that the measurements are heteroscedastic. • Also, the measurements are percentages, which would not be expected to be normally distributed. Example II: Mann-Whiney test (2-tailed test) • The following data provide representative soil moisture contents onsouth and north-facing slopes under grassland in June. The Mann-Whitney test is used to test the null hypothesis that the population medians of the two samples are the same.

  36. Example • A variance ratio test will show that the measurements are heteroscedastic. • Also, the measurements are percentages, which would not be expected to be normally distributed. • U = n1n2 + [n1(n1 + 1)/2] - R1 • U = (14)(17) + [(14)(14 + 1)/2] - 116 = 227 • U0.05(2), 14, 17 = 169 < 227 • p < 0.001, thus reject Ho • The medians of the two samples are significantly different n = 14 R1 = 116 n = 17 R2 = 380

  37. Non-parametric tests for two samples - the Mann-Whitney test for n > 20 • Critical tables for U become unwieldy as n increases above 20 and Mann & Whitney obtained z as a function of U and n as shown below: • z = (U - n1n2/2)/[n1n2(n1 + n2 + 1)/12] • Suppose that the test value of U was found to be 148 with n1 = 16 and n2 = 29. Then • z = (148 - (16  29/ 2)/[(16)(29)(16 + 29 + 1)/12] = -84/42.174 = -1.99 • For a 2-tailed test, the modulus (1.99) is taken. • Referring Table B3, df = infinity, at p = 0.05, z = 1.96 < 1.99 • Thus, reject Ho

  38. Important Notes • Comparisons between two samples can be made with reference to the difference between the sample means or medians • Comparison between sample means are made by calculating their difference and dividing by an appropriate sample standard deviation • Where the sample size is large (e.g. n > 50) a two-sample z test can be used. For small samples, a Student’s t-test can be used • The parametric t and z tests should be applied to independent interval/ ratio measurements which are at least approximately normally distributed • A simple test of homoscedasticity (F ratio test) should be applied prior to application of t and z tests • For non-normally distributed or heteroscedastic data, the non-parametric Mann-Whitney test can be utilized

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