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Chapter 8

Chapter 8. T-tests – comparison of Means. How to find the critical value when alpha, α , is given. Use the t distribution table. If α = .05 and two-tailed test, and the df = 18, what is the critical value?. Ho: There are no differences in the mean scores i.e. µ 1 = µ 2 .

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Chapter 8

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  1. Chapter 8 T-tests – comparison of Means

  2. How to find the critical value when alpha, α, is given • Use the t distribution table. If α = .05 and two-tailed test, and the df = 18, what is the critical value? Ho: There are no differences in the mean scores i.e. µ1 = µ2. H1: µ1≠ µ2 ( a 2-tailed test) meaning µ1 > µ2 or µ1 < µ2 So divide α by 2 .05 /2 = .025 Look at the t distribution table at df = 18 And under α = .025. t critical, t crit = 2.101

  3. p = .025 p = .025 t = + 2.101 t = - 2.101 Reject Null hypothesis Reject Null hypothesis Do not reject Null hypothesis If your calculated t value or t obtained = 3.20, it is in the rejection zone, and so you will reject the null hypothesis and retain the alternative hypothesis of H1 If your calculated t value is = 1.89, it is in the retain (do not reject) zone, so you will not reject the null hypothesis

  4. Use the t distribution table. If α = .05 and one-tailed test, and the df = 18, what is the critical value? Ho: There are no differences in the mean scores i.e. µ1 = µ2. H1: µ1 < µ2 ( a 1-tailed test)

  5. p = ,05 t = + 1.734 Reject Null hypothesis Do not reject Null hypothesis If your calculated t value or t obtained = 3.20, it is in the rejection zone, and so you will reject the null hypothesis and retain the alternative hypothesis of H1 If your calculated t value is = 1.20, it is in the retain (do not reject) zone, so you will not reject the null hypothesis

  6. Type I and Type II Error • Type I error is committed if a true null hypothesis is rejected. • Type II error is committed when a false null hypothesis is retained. TRUE STATE OF AFFAIRS Ho is True Ho is False YOUR DECISION Retain Ho Correct Type II error Reject Ho Type I error Correct

  7. Testing the Significance of the differences between Means Population 1 Population 2 µ1 µ2 σ1 σ2 N1 N2 Sample 1 Sample 2 X1 S1 n1 X2 S2 n2

  8. x1 - x2 |u1 - µ2| Mean of sampling Distribution of x1 - x2

  9. Sampling distribution µ1 - µ2 = 25 µ1 = 25 µ1 = 25 µ1 = 35 µ1 = 27 25 The sampling Distribution of the difference of means |u1 - µ2| = 0 |u1 - µ2| = 10 |u1 - µ2| = 2

  10. The mean of the sampling distribution of the x1-x2 is equal to the difference between the population means u1 - µ2. • If the population is normally distributed, the sampling distribution will be normal. • If sample sizes are sufficiently large, the sampling distribution will be normal whether or not the populations are normally distributed

  11. Assumptions of t-tests • 1) Data must be interval or ratio • 2) Data must be obtained via random sampling from population • 3) Data must be normally distributed • 4) The variance of each group must be similar

  12. Step 1: Define the null and alternative hypotheses: Ho: µ1 = µ2 H1: µ1≠ µ2 Step 2: Set level of significance, alpha. Alpha is set at .05

  13. Step 3: To calculate t: Variance Formula for

  14. Computational Formula for

  15. Example: Test whether there is any significant differences between the Maths marks obtained by Male and Female students Using the computational formula for Male Female 84 88 78 97 67 74 87 80 80 87 78 90 78 90 79 86 82 84 81 78 79.40 - 85.40 t = 63292 – (794)2 73334 – (854)2 10 10 1 + 1 10 10 10 + 10 - 2 t = -2.23 X1 = 79.40 S1 = 5.25 Σx1 = 794 Σx12 = 63292 n1 = 10 X2 = 85.40 S2 = 6.69 Σx2 = 854 Σx22 = 73334 n2 = 10

  16. Step 4: Locate the critical value of t from the t distribution table The degree of freedom is n1 + n2 - 2 10 + 10 – 2 = 18 For the 2-tailed test, alpha for each tail Enter the left column of the t table at 18 and locate the number in the column alpha = .025, the critical value of t is 2.10

  17. Step 5: Compare t obtained of (-2.23) with the critical value of 2.10. Since the absolute value of t obtained is greater than the critical value of 2.10, the null hypothesis is rejected and the alternative hypothesis is retained.

  18. p = .025 p = .025 t = + 2.10 t = - 2.10 Reject Null hypothesis Reject Null hypothesis Do not reject Null hypothesis -2.23 is in the reject zon

  19. Step 6: Interpret and report your findings There are significant differences in the Maths marks between male and female students. The means indicate that female students (x = 85.40) obtain significantly higher Maths marks than male students (x = 79.40).

  20. 4 Assumptions of t-test comparisons • 1) Random sampling: it is assumed that the samples are representative of the populations from which they are drawn. • 2) The populations from which the samples are taken are normally distributed. This assures that the sampling distributions for x1 and x2 are normally distributed. As such the sampling distribution of the differences of the means are also normally distributed. If one of the sample is not normally distributed, the assumption is violated unless the total sample size is sufficiently large (more than 30). The differences of means sampling distribution approach normal distribution as the sample size increases • 3) Homogeneity of variance – the variances of the populations area equal • 4) Independent observations – the scores within each sample are independent of one another – meaning each subject or respondent supplies only one score.

  21. Effect Size for t-test • What is Effect size for a t-test analysis? • It is the standardized measure of the effect of one variable (say Gender) on another variable (say Maths marks).

  22. Effect Size Example: X1 = 15.08 s1 = 4.05 X2 = 14.36 s2 = 3.63 Result: Effect Size (Cohen’s d) = .1875 (Small effect size) Note: Effect size ~ .3 small; ~ .5 (medium); ~ .8 (high)

  23. Effect Size measured by Cohen’s d Cohen’dInterpretation ~ .2 Small ~ .5 Moderate ~ .8 Large

  24. Presentation of t-test results using the APA format Table 2 T-tests comparisons of CRA scores by gender Father Mother Effect Size p < .05 t-value (n =13) (n =12) .18 15.06 14.36 5.38 NS Mean SD 4.05 3.63

  25. Power of a test • Power of a statistical test is the probability of observing a treatment effect when it occurs. • It is the probability that it will correctly lead to the rejection of a false null hypothesis (Green, 2000) • The statistical power is the ability of the test to detect an effect if it actually exists (High, 2000) • The statistical power is denoted by 1 – β, where β is the Type II error, the probability of failing to reject the null hypothesis when it is false. • Conventionally, a test with a power greater than .8 level (or β = < .2) is considered statistically powerful. α = is the probability of rejecting the true null hypothesis (Type I error) β = is the probability of not rejecting the false null hypothesis (Type II error)

  26. There are four components that influence the power of a test: • 1) Sample size, or the number of units (e.g., people) accessible to the study • 2) Effect size, the difference between the means, divided by the standard deviation (i.e. 'sensitivity') • 3) Alpha level (significance level), or the probability that the observed result is due to chance • 4) Power, or the probability that you will observe a treatment effect when it occurs Usually, experimenters can only change the sample size (population) of the study and/or the alpha value

  27. Exercise 1: The following are the Statistics marks obtained for 2 levels of anxiety. • High Anxiety Low Anxiety • x1 = 4.2 x2 = 2.2 • s12 = .5 s22 = .7 • n1 = 10 n2 = 10 • Specify the null and alternative hypothesis • Test the null hypothesis with α = .05. What is the tcrit for α = .05, two-tailed test? • Perform the t test • Interpret the findings • Suppose the variance is increased so that s12=5.2 and s22 = 5.4 • a) perform a t test with α = .05 • b) Interpret the findings. • c) What effect has increasing the variability had on tobt and the final • conclusion about the null hypothesis? • 6) Write your report in the APA format together with the APA table.

  28. Exercise 2: Test whether there is any significant differences between the Motivation levels of Form 5A and 5F students at α = .05. Write your report in the APA format together with the APA table. Form 5A Form 5F 27 24 39 28 25 23 33 25 21 24 35 22 30 29 26 26 25 29 31 28 35 19 30 29 28 X1 = S1 = Σx1 = Σx12 = n1 = X2 = S2 = Σx2 = Σx22 = n2 =

  29. How to test the significance of the differences of means for dependent Samples – Use Paired t-test • Example: You wish to test whether there are significant differences in the Maths scores of a class before and after the treatment or intervention (eg special math tuition) at α = .05 • The same students are involved. • So this comparison is called comparison of the means of dependent samples

  30. Paired t-test • Assumptions 1) Normality of the population difference of scores – this is ascertained by ensuring the normality of each variable separately. 2) the other assumptions similar to group t – test a) Data must be interval or ratio b) Data must be obtained via random sampling from population c) Data must be normally distributed

  31. Example: Maths scores Maths scores D D2 Before tuition after tuition 5 3 2 4 6 4 2 4 7 7 0 0 6 7 -1 1 ΣD2 = 9 ΣD = 3 Step 1: State the null hypothesis and alternate hypothesis Step 2: Compute SD s D =ΣD2 - (ΣD)2 / np np - 1 sD = 9 - (3)2 / 4 = 6.75 = 1.50 4 – 1 3

  32. Step 3: Substitute sD in the following formula: sD = sD np = 1.50 = .75 4 Step 4: Substitute into the following formula to calculate t t = x – y with df = np - 1 sD t = 1.45

  33. Example: t-test comparison for paired data Maths Maths D D2 Pretest Posttest 8 4 4 16 12 8 4 16 6 2 4 6 11 8 15 9 8 5 7 4 .025 .025 tobt =+3.36 0 tcrit = -2.36 tcrit = +2.36 x = 8.88 ΣD = 25 ΣD2= 115 y = 5.75 tobt = 8.88 – 5.75 .81 t obt = 3.86 df = np – 1 T obt = 2.365 Since 3.86 > 2.365, Reject Ho: µx = µy Retain H1: µx ≠ µy SD= SD np = 2.30 8 = .81 sD == ΣD2 - [(ΣD)2/ np] np – 1 = 2.30

  34. Interpretation: Table 1 shows that there is a significant difference between the Maths pretest and Maths posttest mean scores, t (7) = 3.86, p < .05. The means indicate that Maths Posttest mean score are significantly lower (y = 5.75) than the Maths Prestest Mean score (x = 8.88). This indicates that that the treatment (tuition) has a significant negative effect on the students’ performance in Mathematics! Insert your APA Table 1 here

  35. Exercise 3: • The following are the Motivation scores before and after a Motivation • Program. Analyse the data and ascertain whether the Motivation program • Is effective in enhancing the level of motivation of the students. • Pretest Posttest • 80 • 88 70 • 69 88 • 79 89 • 77 87 • 67 56 • 45 67 • 57 81 • 87 67 • 59 67 • State the null and the alternative hypothesis • What is the tobt? • What is the tcrit? • Interprete the findings.

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