# Sd,t test

26 de Aug de 2021                             1 de 29

### Sd,t test

• 2. DIFFERENCE BETWEEN PARAMETRIC AND NON PARAMETRIC STATISTICS PARAMETRIC NON-PARAMETRIC The data should be normally distributed Distribution free(skewed, outliers) Homogeneity of variance Homogeneous or heterogeneous Ratio or interval data Nominal or ordinal data Independent data Any Measure of central tendency - Mean Measure of central tendency – Median Sample size should be large Small sample size Useful in drawing conclusions and generalizations Simple and less effective t-test, ANOVA, Pearson’s correlation, MANOVA, ANCOVA, Regression Chi-square, Mann Whitney U test, Sign test, Kruskal wallis test, Spearman's Rho
• 3. Standard Deviation Measure of variance, SD of a set of scores is defined as the square root of the average of the squares of the deviations of each score from the mean
• 4. Standard Deviation cont… SD is regarded as the most stable and reliable measure of variability It employees mean for its computation Often called root mean square deviation Denoted by Greek letter sigma σ Ungrouped data Grouped data Σ𝒙𝟐/𝑵 Σ𝒇𝒙2/N
• 5. Standard Deviation cont… SD is regarded as the most stable and reliable measure of variability It employees mean for its computation Often called root mean square deviation Denoted by Greek letter sigma σ Ungrouped data Grouped data Σ𝑿𝟐/𝑵 Σ𝒇𝒙2/N
• 6. Calculate the SD for the following data set 52, 50, 56, 68, 65, 62, 57 70, M= 480/8 = 60, N = 8 Scores (x) Deviation from the mean (X – M) or X= 𝑥 − 𝑥 x2 52 -8 (52-60) 64 50 -10 (50-60) 100 56 -4(56-60) 16 68 8 64 65 5 25 62 2 4 57 -3 9 70 10 100 = 382 √382/8 = √47.75 SD = 6.91 𝑥2 𝑁
• 7. Calculate the SD for the following data set 52, 50, 56, 68, 65, 62, 57 70, M= 37.9 , N = 10 Scores (x) De34viation from the mean (X – M) or X= 𝑥 − 𝑥 x2 30 -7.9 62.4 35 2.9 8.4 36 1.9 3.6 39 1.1 1.21 42 4.1 16.8 44 6.1 37.2 46 8.1 65.6 38 0.1 0.01 34 3.9 15.2 35 2.9 8.4 = 219 √219/10 = √21.9 SD = 4.6 𝑥2 𝑁
• 8. Compute SD for the frequency distribution given below IQ scores : 127-129, 124-126, 121-123, 118-120, 115-117, 112-114, 109-111, 106-108, 103-105, 100-102 Frequencies : 1, 2, 3, 1, 6, 4, 3, 2, 1, 1 Mean - 115
• 9. Computation of SD IQ Scores f m(mid point) x=(m-𝑥) x2 fx2 127-129 1 128 13 169 169 124-126 2 125 10 100 200 121-123 3 122 7 49 147 118-120 1 119 4 16 16 115-117 6 116 1 1 6 112-114 4 113 -2 4 16 109-111 3 110 -5 25 75 106-108 2 107 -8 64 128 103-105 1 104 -11 121 121 100-102 1 101 -14 196 196 Mean=115 Σ𝒇𝐱𝟐 = 𝟏𝟎𝟕𝟒 N = sum of all frequencies = 24 Σ𝒇𝒙2/N = √1074/24 √44.74 = 6.69
• 10. Computation of SD IQ Scores f m(mid point) x=(m-𝑥) x2 fx2 21-22 1 21.5 10.4 108.16 108.1 19-20 0 19.5 8.4 70.5 0 17-18 2 17.5 6.4 40. 9 81.8 15-16 2 15.5 4.4 19.3 38.6 13-14 5 13.5 2.4 5.7 28.5 11-12 9 11.5 0.4 0.16 1.44 9-10 4 9.5 -1.6 2.56 10 7-8 3 7.5 -3.6 12.9 38.7 5-6 2 5.5 -5.6 31.3 62.6 3-4 1 3.5 -7.6 57.7 57.7 1-2 1 1.5 -9.6 92.16 92.1 Mean=11.16 Σ𝒇𝐱𝟐 = N = sum of all frequencies = 24 Σ𝒇𝒙2/N = √1074/24 √44.74 = 6.69
• 11. Computation of SD IQ Scores f m(mid point) fm x=(m-𝑥) x2 Fx2/ 80-84 4 82 328 -12 144 576 85-89 4 87 348 -7 49 196 90-94 3 92 276 -2 4 12 95-99 0 97 0 3 9 0 100-104 3 102 306 8 64 192 105-109 3 107 321 13 169 507 110-114 1 112 112 18 324 324 Mean=1691/18=94 Σ𝒇𝐱𝟐 = N = sum of all frequencies = 24 Σ𝒇𝒙2/N = √1807/18 √ = 10
• 12. When to use Standard Deviation When we need a most reliable measure of variability If there is a need of computation of the correlation coefficients, significance of difference between means Measure of central tendency is available in the form of mean The distribution is normal
• 13. Compute SD for the following 1. 30, 35, 36, 39, 42, 44, 46, 38, 34, 35 2. 80-84, 85-89, 90-94, 95-99, 100-104, 105-109, 110-114 ; F – 4, 4, 3, 0, 3, 3, 1 3. Scores : 21-22, 19-20, 17-18, 15-16, 13-14, 11-12, 9-10, 7-8, 5-6, 3-4, 1-2 F – 1, 0, 2, 2, 5, 9, 4, 3, 2, 1, 1
• 14. SIGNIFICANCE OF DIFFERENCE BETWEEN THE MEANS ‘t ‘- TEST The process of determining the difference between two given sample means varies with respect to the size of the group Case 1 – large but independent samples Case 2 – small but independent samples Case 3 – large but correlated samples Case 4 – small but correlated samples
• 15. Paired- X1PRETEST LSCS-------O(ONINE CLASS)----------X2 POST TEST DEGREE OF FREEDOM=n-1 UNPAIRED –XI LSCS-----------O-------------X2 (30) Y1 FTND-----------------O1------------------Y2(30) EFDECCTIVENESS OF WALKING ON INVOLUTION OF UTERUS AMONG LSCS MOTHERS Df=n1+n2-2 30+30-2
• 16. •In all the cases the experimenter has to set up a null hypothesis and take a decision about the level of significance at which he wishes to test his null hypothesis •Then he has to determine the SE of the difference between the two means and compute the Z score or t ratio •Finally the experimenter has to take a decision about the significance of the standard scores at a given level of significance and reject or retain the null hypothesis SIGNIFICANCE OF DIFFERENCE BETWEEN THE MEANS t - TEST
• 17. •Step 1- To determine the SE of the difference between the means of two samples SED or σD = 𝜎1 2 𝑛1 + 𝜎2 2 𝑛2 •Step 2 – To compute the difference in sample means(observed difference) M1-M2 •Step-5: z/t test = M1- M2 / σD (difference between mean divided by standard error of difference between means) SIGNIFICANCE OF DIFFERENCE BETWEEN THE MEANS t - TEST
• 18. Step6 – Testing the null hypothesis- df =( N1+N2 – 2) or (N-1) Refer to table C of t –distribution and heck table value Step-7 : Give the inference If the calculated t value is greater than the table value at the probability level of 0.5 or 0.01or 0.1 then we will fail to accept null hypothesis. SIGNIFICANCE OF DIFFERENCE BETWEEN THE MEANS t - TEST
• 19. PROCEDURE- FOR CASE1&2 Step 3 - Testing the null hypothesis at some pre-established level of significance The null hypothesis - there exists no real difference between the two sample means is tested against its possible rejection at 5% or 1% level of significance in the following manner i) If the computed value is equal to 1.96 or greater than 1.96 then the null hypothesis is rejected at 5% level of significance ii) If the computed value is equal to 2.58 or greater than 2.58 then the null hypothesis is rejected at 1% level of significance iii) If the computed value is greater than 1.96 but less than 2.58 then the null hypothesis is rejected at 5% level of significance but not rejected at 1% level of significance
• 20. ILLUSTRATION A science teacher wanted to know the relative effectiveness of lecture cum demonstration method over the traditional lecture method. He divided his class into two equal random group A and B and taught group A by the lecture cum demonstration method and group B by the lecture method. After teaching for three months he administered an achievement test to both the groups the data collected were as under Group A Group B Mean 43(M1) 30(M2) SD 8(SD1) 7(SD2) No. of students 65(n1) 65(n2)
• 21. Effectiveness of online classes on knowledge of III year degree students MARKS A B C D E F Pretest 2 4 6 8 2 2 Post test 4 6 8 10 16 4 M1=24/6=4 M2=48/6=8 SD1=2.3 SD2=4.6 N1=6 N2=6
• 22. X-XBAR Post test (Y) Y-y bar 2 2-4=-2 4 4 4 0 0 6 6 2 4 8 8 4 16 10 2 -2 4 16 2 -2 4 4 MEAN=4 32 Mean =8 =√32/6=2.3
• 23. CASE 2 SIGNIFICANCE OF THE DIFFERENCE BETWEEN TWO MEANS FOR SMALL BUT INDEPENDENT SAMPLES •Step 1 - Computation of the standard error of the difference between two means •Here SD is pooled for the two samples • σ = ( 𝑥1 2 + 𝑥2 2) _________________ 𝑁1 − 1 + (𝑁2 − 1) •If we are given the values of SD1 and SD2 instead of raw scores then we have to compute the values for 𝑥1 2 and 𝑥2 2 with the help of these formulas • 𝑥1 2 = SD1 2 (N1 - 1) , 𝑥2 2 = SD2 2 (N2 - 1)
• 24. ILLUSTRATION Group 1 – 10, 9, 8, 7, 7, 8, 6, 5, 6, 4 Group 2 – 9,8, 6, 7, 8, 8, 11, 12, 6, 5 Find M1 and M2, find pooled SD, t and df
• 25. ILLUSTRATION X1 M1 x1 x1 2 X2 M2 x2 x2 2 10 7 3 9 9 8 1 1 9 7 2 4 8 8 0 0 8 7 1 1 6 8 -2 4 7 7 0 0 7 8 -1 1 7 7 0 0 8 8 0 0 8 7 1 1 8 8 0 0 6 7 -1 1 11 8 3 9 5 7 -2 4 12 8 4 16 6 7 -1 1 6 8 -2 4 4 7 -3 9 5 8 -3 9
• 26. PRACTICE A language teacher divides the class into two groups experimental group and control group. Under the assumption that newspaper reading will increase the vocabulary the experimental group was given two hours daily to read English newspapers and magazines while no such facility was provided to the control group. After 6 months both the groups were given a vocabulary test and the scores obtained are detailed below Experimental group – 115, 112, 109, 112, 137 Control group – 110, 112, 95, 105, 111, 97, 112,102
• 27. CASE 3 AND 4 SIGNIFICANCE OF THE DIFFERENCE BETWEEN TWO MEANS FOR CORRELATED SAMPLES – LARGE AND SMALL •Single group method – Repetition of a test to a group of subjects called the initial test then the desired experiment (intervention) is given, followed by the same test to the same individuals •The initial and final test data are correlated •σD = √σM1 2+σM2 2 − 2rσM1σM2 •σM1 = Standard error of the initial test, σM2 = Standard error of the final test •r = coefficient of correlation between scores on initial and final testing
• 28. •Equivalent group method – When we have to compare relative effect of one method or treatment over the other with two groups (experimental and control) •Matching pair technique – Matching is done before the initial test in which each individual is paired with an equivalent match in the other group in terms of variables that are going to affect the results of the study such as age, socio economic background, intelligence etc. •Matching group technique – the group as a whole is matched with another group in terms of their mean, SD. CASE 3 AND 4 SIGNIFICANCE OF THE DIFFERENCE BETWEEN TWO MEANS FOR CORRELATED SAMPLES – LARGE AND SMALL
• 29. In case of grouped matched data the formula is σD = √(σM1 2+σM2 2 ) (1−r2) Illustration : Initial test mean = 70, SD = 6 Final test mean = 67,SD = 5.8 r = 0.82 CASE 3 AND 4 SIGNIFICANCE OF THE DIFFERENCE BETWEEN TWO MEANS FOR CORRELATED SAMPLES – LARGE AND SMALL