T-TEST FOR HYPOTHESIS ABOUT

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1 T-TEST FOR HYPOTHESIS ABOUT Previously we tested the hypothesis that a sample comes from a population with a specified using the normal distribution and a z-test. But the z-test required the population standard deviation (i.e., ). Specifically, The statistic z calculated in this way follows a normal distribution, with = 0 and = 1, which allows us to calculate probabilities of different values for z given the H0 is true. But obviously the formula cannot be used if is not known. If is not known, we can still obtain an estimate of the population standard deviation from the sample; that is, we can calculate s, the sample standard deviation, and substitute s for in the above formula. This produces a statistic known as t (or sometimes Students t). The substitution of s for has, or at least can have, a profound effect on the distribution of the statistic. Specifically, the t distribution will have more variability associated with it than the z distribution; that is, the standard deviation of t will be greater than 1, the standard deviation for z. We will demonstrate this shortly, but that t must have more variability can be appreciated by thinking of the sources of variation from sample to sample in the two formula. The only quantity that changes from sample to sample with z is the sample mean. This is therefore the sole source of variability in z. But for t, both and s will vary from sample to sample. Sometimes, s will be small, which will produce larger values for t than for z, and sometimes s will be large, which will produce smaller values for t than for z. The additional source of variable, s, then, leads to greater variability in t than z; that is, t > z = 1. This means that we cannot use the normal distribution, which requires = 1. We can demonstrate this property of t using the 50,000 samples generated in the last unit to illustrate the sampling distribution of. Recall that the 50,000 samples were selected randomly from a population with = 100 and = 15. First, we calculate z for each sample using these values and the above formula. The 50,000 zs are plotted in the histogram to the right. Note that the mean of the zs is very close to 0 and the standard deviation of the zs is very close to 1, exactly what we expect from the Central Limit Theorem. Also, the distribution is very close to normal (the dark line). Next we calculate 50,000 ts using s

2 instead of in the denominator. Because s varies from sample to sample, there is more variation in the ts than in the zs, as shown in the second histogram. One indicator of the greater variability is the standard deviation of the ts, which is 1.15, much greater than the standard deviation of the zs. A second indicator is the disparity between the frequency distribution for the ts and the fit of the normal distribution (the solid line). There are too few observations near the center and far too many at the extremes. Although the variability is wrong for the normal distribution, note that the mean of the 50,000 ts is still very close to 0 ( ) To illustrate even more concretely the greater variability in t than z, the first few samples are shown below, along with the calculated values for t and z. Note that when s (SD below) is less than = 15, t is greater than z, and that when s is greater than = 15, t is less than z. x1 x2 x3 x4 x5 x6 x7 x8 x9 Mean SD z t The disparity between the distributions for z and t will depend in part on the sample size, n. When n is very large, s will not vary a great deal from sample to sample, and will generally be quite close to ; therefore, the z and t distributions will be quite similar. When n is small, however, s will vary more markedly from sample to sample, and sometimes will be quite far from. Because of this, the probability distribution for t depends in part on the number of observations, or more correctly, the degrees of freedom associated with the t statistic. For the test of a single population

3 mean, the degrees of freedom (df) are obtained from the df for the s that appears in the denominator. This is n - 1. Figure 13.1 from Pagano, demonstrates how the distribution of t changes as df increases from 1 to 5 to 20 to infinity (equals normal distribution). The figure is reproduced to the right. Note that values of t in the tails (ends) of the distribution become more likely with smaller df (i.e., n). This means that we cannot use a single critical value to reject H0 no matter what the df for t, but rather must select critical values separately for studies involving different df. Studies with few df will have more extreme cut-off or critical values for t, whereas studies with more df will have less extreme critical values. As df increases, the critical values for t approach those of critical values for z (e.g., for alpha =.05 and directional test), ultimately reaching the same values as z for df = infinity. We can use the zs and ts calculated for our 50,000 samples to illustrate the issue (see earlier table for a selection of these statistics. Of the 50,000 zs, approximately 4.9% are greater than 1.645, very close to the 5.0% expected for the normal distribution. But 7.0% of the ts are greater than 1.645, far higher than the probability for z. Appendix D in Pagano provides values for t given varying df. For the single sample t-test discussed here, df = n - 1. It is important to keep in mind, however, that there are various uses of t and each will have the appropriate df; for example, the t for the significance of the correlation coefficient will have df = n - 2. Appendix D is reproduced at the end of this handout. Only certain commonly used values for alpha are presented in the table. The appropriate column to use will depend on whether the research involves a directional (one-tailed) or nondirectional (two-tailed) test and the value chosen for alpha (the probability of a Type I error). The row will be determined by df = n - 1 for the single sample t-test. Some probabilities from Appendix D are shown below, along with corresponding values for z. Note that the critical values for z do not change with the df, whereas the critical values for t get smaller as df increases. If instead of 1.645, the critical value for z, we calculate the number of our 50,000 ts greater than (obtained from the table for t using df = 9, our sample size), then 5.0% of the ts are greater than 1.860, equal to the desired value of.05.

4 We can now proceed with our re-analysis of the UofW IQ data using the t-test instead of the z-test. Calculations of descriptive statistics for the UofW Students IQ Problem are shown below. S X X = = 980/9 = X 2 = 107, SSx = X 2 - (X) 2 /n = 107, /9 = = (X - ) s x = [ /(9-1)] = The hypothesis testing procedure follows along the basic plan seen previously for the binomial and the z test. Our null and alternative hypotheses are in fact the same as for the z-test. H0: = 100 Ha: > 100 df = n - 1 = 8 =.05 Directional Ha t Critical = t Observed = ( ) / (12.252/9) = 8.89 / = > t Critical = 1.86 Reject H0 The critical value of t is obtained from df = 8 row and.05 (one-tailed) column. One-Tailed Probability z t df Given the directional alternative hypothesis, the researchers would conclude to Reject Ho: = 100 and Accept Ha: >100. Note that the conclusion would have been different if a nondirectional alternative hypothesis had been used. In that case, the researchers must consider the possibility that IQs for UofW students can be either greater or lower than 100. They must therefore include negative values for the t; that is, they must allocate one-half of alpha (i.e.,.025 assuming alpha =.05) to the lower tail of the distribution and the other half to the upper tail of the distribution. The t Critical would be and t Observed = is less than t Critical. Therefore, H0 is not rejected.

5 SINGLE SAMPLE PROBLEMS (T AND Z) 1. Historical records indicate that police in Winnipeg give out an average of tickets a day, SD = To determine whether more tickets are given out on holiday Mondays during the summer, a random sample of 12 such days was selected. The mean for the 12 days was , SD = What conclusions are warranted? What conclusions would be drawn if the long-term SD was not known and the sample SD had to be used? 2. People with back-pain normally miss days of work per year. A Human Resources psychologist initiated a relaxation and cognitive change program for 8 people with back-pain. The following year, they missed the following number of days of work: 20, 19, 27, 22, 23, 18, 18, and 21. Does the program appear to be of benefit? 3. University students often report considerable stress during their first year, with the average score on the Winnipeg University Stress Scale (WUSS) being 32.90, SD = The Director of Student Services pilot tested a new orientation program to alleviate the stress. The 16 students exposed to the program obtained a mean WUSS score of Would you recommend using the new orientation program (it is expensive, so the University wants to be certain that it works and decides to use alpha =.01). 4. A random sample of 21 media people attending the Fringe Festival were asked to indicate their satisfaction with the shows that they had seen, with -5 being Much Worse than previous years, 0 being the Same as previous years, and +5 being Much Better than previous years. The average rating of the 21 respondents was.56, SD = Do the results provide any evidence that this year s shows were better or worse than previous years? 5. Educational psychologists evaluated a new program to teach basic number skills to young children. In the past, grade 1 children s average score on the Winnipeg Numeracy Scale (WNS) has been A random sample of 6 children exposed to the new program produced the following WNS scores: 75, 77, 82, 70, 71, and 85. Does the program appear to be effective? 6. A coach has been having a lot of problems with his team s foul shooting. The percentage for his top 5 players is only 55% (i.e., these players only make 55% of all foul shots that they take). A Sports Psychologist worked with the team for a week, teaching them visualization techniques and ways to cope with distraction. In the following 4 games, the 5 players make 56, 58, 63, 66, and 54 per cent of their foul shots. Does the program appear to be effective? Because the program is cheap to implement, the coach has decided to use a.10 level of significance. 7. Participants in Alcoholics Anonymous programs who were followed up 2 years after starting the program tended to remain sober for an average of days (no better than people who try to control their drinking on their own). A progressive town hires a psychologist to implement a more effective Cognitive-Behavioural program. The first 15 participants in the program were followed up two-years after starting, and had remained sober for an average of days, SD = days. Does the new program appear to be worth the investment in a professional psychologist?

6 SINGLE SAMPLE T/Z SOLUTIONS 1. Ho: = , Accept Ha > Using = z = ( )/34.56/12 = 13.77/ = =.05, z crit = Do not rej Ho Using s = t = ( )/23.45/12 = 13.77/6.769 = =.05, df = 12-1 = 11, t crit = Rej Ho, Acc Ha, more tickets on holiday Mondays 2. Ho: = Ha: < M = 21.00, SD = t = ( ) / (3.024/8) = -2.76/1.069 = =.05, one-tailed, df = 8-1 = 7, t crit = Rej Ho, Acc Ha, program effective at reducing days missed 3. Ho: = 32.90, Ha: < = z = ( )/ = -8.67/2.82 = =.01, one-tailed, z crit = Rej Ho, Acc Ha, orientation reduces stress 4. Ho: = 0, Ha: =/ 0 t = (.56-0)/1.31/21 =.56/.286 = =.05, df = 21-1 = 20, t crit = Do not rej Ho [95% Conf Int =.56 ± x 1.31/21 =.56 ±.5963 = to 1.156] 5. Ho: = 73.4 Ha: > 73.4 M = , SD = t = ( ) / (5.955/6) = / = =.05, one-tailed, df = 6-1 = 5, t crit = Do Not Rej Ho 6. Ho: = 55 Ha: > 55 M = 59.4, SD = 4.98 t = ( ) / (4.98/5) = 4.4/2.227 = =.10, one-tailed, df = 5-1 = 4, t crit = Rej Ho, Acc Ha, program is effective 7. Ho: mu = Ha: mu > t = ( ) / (45.23/sqrt(15)) = / = 6.24 df = 15-1 = 14, t critical = 1.761, Reject Ho