- 1. t-test .Mahmoud Alhussami, DSc., Ph.D
- 2. Learning Objectives • Compute by hand and interpret – Single sample t – Independent samples t – Dependent samples t • Use SPSS to compute the same tests and interpret the output
- 3. Review 6 Steps for Significance Testing 1. Set alpha (p 4. Find the critical level). value of the 2. State hypotheses, statistic. Null and 5. State the decision Alternative. rule. 3. Calculate the test 6. State the statistic (sample conclusion. value).
- 4. t-test • t –test is about means: distribution and evaluation for group distribution • Withdrawn form the normal distribution • The shape of distribution depend on sample size and, the sum of all distributions is a normal distribution • t- distribution is based on sample size and vary according to the degrees of freedom
- 5. What is the t -test • t test is a useful technique for comparing mean values of two sets of numbers. • The comparison will provide you with a statistic for evaluating whether the difference between two means is statistically significant. • t test can be used either : 1.to compare two independent groups (independent- samples t test) 2.to compare observations from two measurement occasions for the same group (paired-samples t test).
- 6. What is the t -test • The null hypothesis states that any difference between the two means is a result to difference in distribution. • Remember, both samples drawn randomly form the same population. • Comparing the chance of having difference is one group due to difference in distribution. • Assuming that both distributions came from the same population, both distribution has to be equal.
- 7. What is the t -test • Then, what we intend: “To find the he difference due to chance” • Logically, The larger the difference in means, the more likely to find a significant t test. • But, recall: 1. Variability More (less) variability = less overlap = larger difference 2. Sample size Larger sample size = less variability (pop) = larger difference
- 8. Types 1. The one-sample t test is used compare a single sample with a population value. For example, a test could be conducted to compare the average salary of nurses within a company with a value that was known to represent the national average for nurses. 2. The independent-sample t test is used to compare two groups' scores on the same variable. For example, it could be used to compare the salaries of nurses and physicians to evaluate whether there is a difference in their salaries. 3. The paired-sample t test is used to compare the means of two variables within a single group. For example, it could be used to see if there is a statistically significant difference between starting salaries and current salaries among the general nurses in an organization.
- 9. Assumptions of t-Test • Dependent variables are interval or ratio. • The population from which samples are drawn is normally distributed. • Samples are randomly selected. • The groups have equal variance (Homogeneity of variance). • The t-statistic is robust (it is reasonably reliable even if assumptions are not fully met.
- 10. Assumption 1. Should be continuous (I/R) 2. the groups should be randomly drawn from normally distributed and independent populations e.g. Male X Female Nurse X Physician Manager X Staff NO OVER LAP
- 11. Assumption 3. the independent variable is categorical with two levels 4. Distribution for the two independent variables is normal 5. Equal variance (homogeneity of variance) 6. large variation = less likely to have sig t test = accepting null hypothesis (fail to reject) = Type II error = a threat to power Sending an innocent to jail for no significant reason
- 12. Story of power and sample size • Power is the probability of rejecting the null hypothesis • The larger the sample size is most probability to be closer to population distribution • Therefore, the sample and pop distribution will have less variation • Less variation the more likely to reject the null hypothesis • So, larger sample size = more power = significant t test
- 13. (One Sample Exercise (1 Testing whether light bulbs have a life of 1000 hours 1. Set alpha. α = .05 2. State hypotheses. – Null hypothesis is H0: µ = 1000. – Alternative hypothesis is H1: µ ≠ 1000. 3. Calculate the test statistic
- 14. Calculating the Single Sample t 800 What is the mean of our sample? 750 X = 867 940 What is the standard deviation 970 for our sample of light bulbs? 790 SD= 96.73 980 SD 96.73 SE = = = 30.59 820 N 10 760 X − µ 867 − 1000 1000 tX = = = −4.35 860 SX 30.59
- 15. Determining Significance 4. Determine the critical value. Look up in the table (Heiman, p. 708). Looking for alpha = .05, two tails with df = 10-1 = 9. Table says 2.262. 5. State decision rule. If absolute value of sample is greater than critical value, reject null. If |-4.35| > |2.262|, reject H0.
- 16. Finding Critical Values A portion of the t distribution table
- 17. t Values • Critical value decreases if N is increased. • Critical value decreases if alpha is increased. • Differences between the means will not have to be as large to find sig if N is large or alpha is
- 18. Stating the Conclusion 6. State the conclusion. We reject the null hypothesis that the bulbs were drawn from a population in which the average life is 1000 hrs. The difference between our sample mean (867) and the mean of the population (1000) is SO different that it is unlikely that our sample could have been drawn from a population with an average life of 1000 hours.
- 19. SPSS Results One-Sample Statistics Std. Error N Mean Std. Deviation Mean BULBLIFE 10 867.0000 96.7299 30.5887 One-Sample Test Test Value = 1000 95% Confidence Interval of the Mean Difference t df Sig. (2-tailed) Difference Lower Upper BULBLIFE -4.348 9 .002 -133.0000 -202.1964 -63.8036 Computers print p values rather than critical values. If p (Sig.) is less than .05, it’s significant.
- 20. Steps For Comparing Groups
- 21. t-tests with Two Samples Independent Samples t-test Dependent Samples t-test
- 22. Independent Samples t-test • Used when we have two independent samples, e.g., treatment and control groups. X1 − X 2 • Formula is: t X1 − X 2 = SEdiff • Terms in the numerator are the sample means. • Term in the denominator is the standard error of the difference between means.
- 23. Independent samples t-test The formula for the standard error of the difference in means: 2 2 SD1 SD2 SEdiff = + N1 N2 Suppose we study the effect of caffeine on a motor test where the task is to keep a the mouse centered on a moving dot. Everyone gets a drink; half get caffeine, half get placebo; nobody knows who got what.
- 24. Independent Sample Data )(Data are time off task )Experimental (Caff Control (No Caffeine) 12 21 14 18 10 14 8 20 16 11 5 19 3 8 9 12 11 13 15 N1=9, M1=9.778, SD1=4.1164 N2=10, M2=15.1, SD2=4.2805
- 25. Independent Sample )Steps(1 1. Set alpha. Alpha = .05 2. State Hypotheses. Null is H0: µ1 = µ2. Alternative is H1: µ1 ≠ µ2.
- 26. Independent Sample )Steps(2 3. Calculate test statistic: X 1 − X 2 9.778 − 15.1 − 5.322 t= = = = −2.758 SEdiff 1.93 1.93 2 2 2 2 SD SD (4.1164) (4.2805) SEdiff = +1 = + 2 = 1.93 N1 N2 9 10
- 27. Independent Sample )Steps(2 3. Calculate test statistic: X 1 − X 2 9.778 − 15.1 − 5.322 t= = = = −2.758 SEdiff 1.93 1.93 2 2 2 2 SD SD (4.1164) (4.2805) SEdiff = +1 = + 2 = 1.93 N1 N2 9 10
- 28. Independent Sample Steps )(3 4. Determine the critical value. Alpha is . 05, 2 tails, and df = N1+N2-2 or 10+9-2 = 17. The value is 2.11. 5. State decision rule. If |-2.758| > 2.11, then reject the null. 6. Conclusion: Reject the null. the population means are different. Caffeine has an effect on the motor pursuit task.
- 29. Using SPSS • Open SPSS • Open file “SPSS Examples” for Lab 5 • Go to: – “Analyze” then “Compare Means” – Choose “Independent samples t-test” – Put IV in “grouping variable” and DV in “test variable” box. – Define grouping variable numbers. • E.g., we labeled the experimental group as “1” in our data set and the control group as “2”
- 30. Independent Samples Exercise Experimental Control 12 20 14 18 10 14 8 20 16 Work this problem by hand and with SPSS. You will have to enter the data into SPSS.
- 31. SPSS Results Group Statistics Std. Error GROUP N Mean Std. Deviation Mean TIME experimental group 5 12.0000 3.1623 1.4142 control group 4 18.0000 2.8284 1.4142 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Mean Std. Error Difference F Sig. t df Sig. (2-tailed) Difference Difference Lower Upper TIME Equal variances .130 .729 -2.958 7 .021 -6.0000 2.0284 -10.7963 -1.2037 assumed Equal variances -3.000 6.857 .020 -6.0000 2.0000 -10.7493 -1.2507 not assumed
- 33. Dependent Samples t-test • Used when we have dependent samples – matched, paired or tied somehow – Repeated measures – Brother & sister, husband & wife – Left hand, right hand, etc. • Useful to control individual differences. Can result in more powerful test than independent samples t-test.
- 34. Dependent Samples t Formulas: D tXD = SEdiff t is the difference in means over a standard error. SDD SEdiff = n pairs The standard error is found by finding the difference between each pair of observations. The standard deviation of these difference is SDD. Divide SDD by sqrt(number of pairs) to get SEdiff.
- 35. Another way to write the formula D tXD = SDD n pairs
- 36. Dependent Samples t example Person Painfree Placebo Difference (time in sec) 1 60 55 5 2 35 20 15 3 70 60 10 4 50 45 5 5 60 60 0 M 55 48 7 SD 13.23 16.81 5.70
- 37. Dependent Samples t )Example (2 1. Set alpha = .05 2. Null hypothesis: H0: µ1 = µ2. Alternative is H1: µ1 ≠ µ2. 3. Calculate the test statistic: SD 5.70 SEdiff = = = 2.55 n pairs 5 D 55 − 48 7 t= = = = 2.75 SEdiff 2.55 2.55
- 38. Dependent Samples t )Example (3 4. Determine the critical value of t. Alpha =.05, tails=2 df = N(pairs)-1 =5-1=4. Critical value is 2.776 5. Decision rule: is absolute value of sample value larger than critical value? 6. Conclusion. Not (quite) significant. Painfree does not have an effect.
- 39. Using SPSS for dependent t- test • Open SPSS • Open file “SPSS Examples” (same as before) • Go to: – “Analyze” then “Compare Means” – Choose “Paired samples t-test” – Choose the two IV conditions you are comparing. Put in “paired variables box.”
- 40. Dependent t- SPSS output Paired Samples Statistics Std. Error Mean N Std. Deviation Mean Pair PAINFREE 55.0000 5 13.2288 5.9161 1 PLACEBO 48.0000 5 16.8077 7.5166 Paired Samples Correlations N Correlation Sig. Pair 1 PAINFREE & PLACEBO 5 .956 .011 Paired Samples Test Paired Differences 95% Confidence Interval of the Std. Error Difference Mean Std. Deviation Mean Lower Upper t df Sig. (2-tailed) Pair 1 PAINFREE - PLACEBO 7.0000 5.7009 2.5495 -7.86E-02 14.0786 2.746 4 .052
- 41. Relationship between t Statistic and Power • To increase power: – Increase the difference between the means. – Reduce the variance – Increase N – Increase α from α = . 01 to α = .05
- 42. To Increase Power • Increase alpha, Power for α = .10 is greater than power for α = .05 • Increase the difference between means. • Decrease the sd’s of the groups. • Increase N.
- 43. Calculation of Power From Table A.1 Zβ of . 54 is 20.5% Power is 20.5% + 50% = 70.5% In this example Power (1 - β ) = 70.5%
- 44. Calculation of Sample Size to Produce a Given Power Compute Sample Size N for a Power of .80 at p = 0.05 The area of Zβ must be 30% (50% + 30% = 80%) From Table A.1 Zβ = .84 If the Mean Difference is 5 and SD is 6 then 22.6 subjects would be required to have a power of .80
- 45. Power • Research performed with insufficient power may result in a Type II error, • Or waste time and money on a study that has little chance of rejecting the null. • In power calculation, the values for mean and sd are usually not known beforehand. • Either do a PILOT study or use prior research on similar subjects to estimate the mean and sd.
- 46. Independent t-Test For an Independent t-Test you need a grouping variable to define the groups. In this case the variable Group is defined as 1 = Active 2 = Passive Use value labels in SPSS
- 47. Independent t-Test: Defining Variables Be sure to enter value Grouping variable GROUP, the level of labels. measurement is Nominal.
- 49. Independent t-Test: Independent & Dependent Variables
- 50. Independent t-Test: Define Groups
- 52. Group Statistics Independent t-Test: Group N Mean Std. Deviation Std. Error Mean Output Ab_Error Active 10 2.2820 1.24438 .39351 Passive 10 1.9660 1.50606 .47626 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means 95% Confidence Interval of the Mean Std. Error Difference F Sig. t df Sig. (2-tailed) Difference Difference Lower Upper Ab_Error Equal variances .513 .483 .511 18 .615 .31600 .61780 -.98194 1.61394 assumed Equal variances .511 17.382 .615 .31600 .61780 -.98526 1.61726 not assumed Assumptions: Groups have equal variance [F = .513, p =.483, YOU DO NOT WANT THIS TO Are the groups BE SIGNIFICANT. The groups have equal different? variance, you have not violated an assumption of t-statistic. t(18) = .511, p = .615 NO DIFFERENCE 2.28 is not different from 1.96
- 53. Dependent or Paired t-Test: Define Variables
- 54. Dependent or Paired t-Test: Select Paired-Samples
- 55. Dependent or Paired t-Test: Select Variables
- 56. Dependent or Paired t-Test: Options
- 57. Paired Samples Statistics Std. Error Dependent or Paired Pair Pre Mean 4.7000 N 10 Std. Deviation 2.11082 Mean .66750 t-Test: Output 1 Post 6.2000 10 2.85968 .90431 Paired Samples Correlations N Correlation Sig. Pair 1 Pre & Post 10 .968 .000 Paired Samples Test Paired Differences 95% Confidence Interval of the Std. Error Difference Mean Std. Deviation Mean Lower Upper t df Sig. (2-tailed) Pair 1 Pre - Post -1.50000 .97183 .30732 -2.19520 -.80480 -4.881 9 .001 Is there a difference between pre & post? t(9) = -4.881, p = .001 Yes, 4.7 is significantly different from 6.2

- 1 . Set Alpha level, probability of Type I error, that is probability that we will conclude there is a difference when there really is not. Typically set at .05, or 5 chances in 100 to be wrong in this way . 2 . State hypotheses. Null hypothesis: represents a position that the treatment has no effect. Alternative hypothesis is that the treatment has an effect. In the light bulb example Ho: mu = 1000 hours H1: mu is not equal to 1000 hours 3 . Calculate the test statistic. (see next slide for values ) 4 . Determine the critical value of the statistic . 5 . State the decision rule: e.g., if the statistic computed is greater than the critical value, then reject the null hypothesis . Conclusion: the result is significant or it is not significant. Write up the results .
- Let’s do the steps : 1 . Set alpha = .05. If there is no difference, we will be wrong only 5 times in 100 . 2 . State hypotheses. (Null) H 0 : = 1000. (Alternative) H 1 : 1000. We are testing to see if our light bulbs came from a population where average life is 1000 hours . 3 . Calculate the test statistic .
- Go over the answers to exercise with them . M = 867 SD = 96.7299 SE= 30.58867 t = -4.35 Reject H 0 Bulbs were not drawn from population with 1000 hr life . Any questions ?
- 4 . Determine the critical value of the statistic. We look this up in a table. We need to know alpha (.05, two-tailed) and the degrees of freedom (df). For this test, the df are N-1, in our case 10-1 = 9. According to the table, the critical value is 2.262 . 5 . State the decision rule: If the absolute value of the test statistic is greater that the critical value, we reject the null hypothesis. In our case, if |-.33| is greater than 2.262, we reject the hypothesis that = 1000. This is not the case here .
- State the conclusion . Our results suggest that GE’s claim that their light bulbs last 1,000 hours is FALSE. (because we had a sample of 10 GE light bulbs and our sample mean was so far away from 1,000 hours that it is highly unlikely that these bulbs came from a population of bulbs whose mean is really 1,000.) There is a 5% chance that this conclusion is wrong (I.e., we may have gotten a difference this big just by chance factors alone ).
- On Brannick’s website, Research Methods, Labs, Lab Presentations. Click on Lab 5 SPSS Examples, the Open. SPSS should run and the data for this lab should appear. In the middle is the column ltbulb. This has the data for the ligtbulb example. In the SPSS data editor, click Analyze, Compare Means, One-Sample T Test. Select ltbulb and put it in the Test Variables box. Type 1000 in the Test Value box. Click OK. You get the output on this slide .
- Here we have two different samples, and we want to know if they were drawn from populations with two different means. This is equivalent to saying whether a treatment has an effect given a treatment group and a control group. The formula for this t is on the slide . Here t is the test statistic, and the terms in the numerator are the sample and population means. The term in the denominator is SE diff , which is the standard error of the difference between means . You can see from the subscripts for both t and SE, that we are now dealing with the Sampling Distribution of the DIFFERENCE between the means. This is very similar to the sampling distribution that we created last week. However, what we would do to create a sampling distribution of the differences between the means is rather than selecting 5 scores and computing a mean, we would select 5 pairs of scores, subtract one value from the other, then calculate the mean DIFFERENCE value . If we are doing a study and have two groups, what do we EXPECT that the difference in their mean scores will be ? [ They should say zero ]. Thus, the mean of the sampling distribution of the differences between the means is zero . The subscripts are here to tell you which Sampling Distribution we are dealing with (for the Sampling Distribution of Means last week, we had a subscript X-bar. For the sampling distribution of the differences between the means, we have a notation specifying a difference, specifically, the difference between X-bar1 and X-bar2 .
- Suppose we have two samples taken from the same population. Suppose we compute the mean for each sample and subtract the mean for sample 2 from sample 1. We will get a difference between sample means. If we do this a lot, on average, that difference will be zero. Most of the time it won’t be exactly zero, however. The amount that the difference wanders from zero on average is , the standard error of the difference .
- So let’s say we do the following study. We bring in our volunteers and give each of them a psychomotor test where they use a mouse to keep a dot centered on a computer screen target that keeps moving away (pursuit task). One hour before the test, both groups get an oral dose of a drug. For every other person (1/2 of the people), the drug is caffeine. For the other half, it’s a placebo. Nobody in the study knows who got what. All take the test. The results are in the slide .
- 1 . Set alpha = .05, two tailed (just a difference, not a prediction of greater n or less than ). 2 . Null Hypothesis: . This is the same as . This says that there is no difference between the drug group and the placebo group in psychomotor performance in the population. The alternative hypothesis is that the drug does have an effect, or
- 3 . Calculate the test statistic (see the slide ).
- 3 . Calculate the test statistic (see the slide ).
- 4 . Determine the critical value of the statistic. We look this up in a table . Alpha is .05, t is 2-tailed and the df are n 1 +n 2 -2, or in our case, 17. The critical value is 2.110 . 5 . State the decision rule. If the absolute value of the test statistic is larger than the critical value, reject the null hypothesis. If |-2.758| > 2.110, reject the null . 6 . Conclusion, the population means are different. The result is significant at p < .05 .
- Make sure that they look at the data in SPSS to see how the groups were defined and how that relates to the “define groups” task .
- Have them work this one. Assume again that this is time off task for the DV . Here are the answers for the independent samples exercise M1 = 12 , M2 = 18 SD1 = 3.162278, SD2 = 2.8284227 Std Error = 2 t = -6/2 = -3; df = 5 + 4 - 2 = 7 t(.05) = 2.3646; 3 > 2.3646 Reject null hypothesis We conclude that caffeine has an effect . Be sure to cover the relevant areas of the SPSS printout. You should show them where everything that they calculate by hand is on the printout. Also cover the Levene’s test. Explain that if the Levene’s test is significant, we need to use the row that says ‘equal variances NOT assumed”. We do NOT want the Levene’s test to be significant as it violates an assumption of the t-test .
- We use this when we have measures on the same people in both conditions (or other dependency in the data). Usually there are individual differences among people that are relatively enduring. For example, suppose we tested the same people on the psychomotor test twice. Some people would be very good at it. Others would be relatively poor at it. The dependent t allows us to take these individual differences into account. The scores on the variable in one treatment will be correlated with the scores on the other treatment . If the observations are positively correlated (most people score either high on both or low on both) and if there is a difference in means, we are more likely to show it with the dependent t-test than with the independent samples t-test. [Emphasize this point, they need to know it for their homework .]
- We are still dealing with the Sampling Distribution of the Difference between the means. Our subscript is different here, but says basically the same thing. We are looking at the MEAN DIFFERENCE SCORE. The subscript for the independent samples t said we were looking at the DIFFERENCE BETWEEN THE MEANS .
- In this formula, we just put the formula for Se diff in the denominator instead of having you calculate it separately. [this is the formula that appears on the “Guide to Statistics” sheet they can download .
- Suppose that we are testing Painfree, a drug to replace aspirin. Five people are selected to test the drug. On day one, ½ get painfree, and the other get a placebo. Then all put their hands into icewater until it hurts so bad they have to pull their hands from the water. We record how long it takes. The next day, they come back and take the other treatment. (Counterbalancing & double blind .)
- 1 . Set alpha = .05, two tailed (just a difference, not a prediction of greater or less than ). 2 . Null Hypothesis: . This is the same as . This says that there is no difference between the pain killer and the placebo in the population. The alternative hypothesis is that the pain killer does have an effect, or 3 . Calculate the test statistic (see slide ).
- 4 . Determine the critical value of the statistic. We look this up in a table. Alpha is .05, t is 2-tailed and our df are N-1, where N is the number of pairs . In this case df = 5-1 = 4. The critical value is 2.776 . 5 . State the decision rule. If the absolute value of the test statistic is larger than the critical value, reject the null hypothesis. If |2.75| > 2.776, reject the null . 6 . Conclusion. he population means are not (quite) different. The result is not significant at p < .05 .
- Point out that the data for an independent t-test and dependent t-test must be entered differently in SPSS . [ They should choose “painfree” and “placebo” to put in the paired variables box .]
- Go over output. Have them start on their homework or project .