A rope tied to a body is pulled, causing the body to accelerate. But according to Newtons third law, the body pulls back on the rope with a force of equal magnitude and opposite direction. Is the total work done then zero? If so, how can the bodys kinetic energy change? Explain.
Statistics for the Behavioral Sciences EXAM 2 REVIEW Material Covered: 1. Computations (for One and Two sample group tests) 2. APA Format write ups 3. Reading SPSS Output/Levene’s Test 4. Power (and computations) 5. Relationships between Delta, Effect Size, Alpha, Beta, and Power 6. Hypotheses/Type 1 and 2 Errors 7. Significance Interpretation HYPOTHESIS TESTING: Alpha: the amount of error we’re willing to make. It is a specific pvalue (usually 0.05 or 0.01) that we compare to the actual pvalue that we find on the table. Later, this helps us determine significance of our results. The Null Hypothesis (H0) When we assume that the population mean (mu) is equal to the sample mean (mu0). This is the given; this is the hypothesis that the researcher often hopes to disprove. The Alternative/Research Hypothesis (Ha) The population mean (mu) is different from our sample mean (mu0). This is the hypothesis we’re usually interested in – the one we hope yields significant/reliable results because it discovers a new finding. Determining Errors: In real life scenarios, Type 2 errors are more dangerous to make because it’s worse to say “Oh, there’s no significance here” when there really was something significant. What we SHOULD HAVE done (reality) What we CHOSE to do Fail to Reject Reject the Null (below ) Fail to Reject Null Correct Conclusion Type 2 Error Reject Null Type 1 Error Correct Conclusion ONE GROUP T TEST (COMPUTATION) A researcher compares a population of college students with a test score mean of 40 to a sample of 9 NYU college students with a mean of 44 and a sample standard deviation (s) of 5. Is this difference significant in a one sample test with alpha = .05, twotailed 1. Hypotheses: Null: mu = 40 The test score mean of the sample students is the same as the mean of the general population. Alternative: mu ≠ 40 The test score mean of the sample students is not the same as the mean of the general population. 2. Calculating STANDARD ERROR: (This will come in handy towards the end of the problem) We are given the sample standard deviation, so plug that into the formula. We get 1.667. 3. Decide on the appropriate test: z or t If our sample size > 40, we use a ztest. If our sample size < 40, we use a ttest. In this example, our sample size is < 40, so we use the ttest formula. After using this formula, we get a tcalculated score of 2.40. To find tcritical, we match up alpha (0.05) with the df (8) on table A.2. From the table, our tcritical is 2.306. Note: For onesample tests, df is simply (N1). 4. Compare tcalc with tcrit: If tcalc > tcrit, then we reject the null. If tcalc < tcrit then we fail to reject the null. Our tcalc (2.40) > tcrit (2.306), so we reject the null. Because we’re rejecting, we risk making a Type 1 Error. APA WRITE UP: We reject the null hypothesis. The NYU college students scored on average higher (M = 44, s = 5) than the general population (μ = 40); a onesample t test with alpha = 0.05 demonstrates that this difference is significant, t(8) = 2.40, p < .05, two tailed. 5. Confidence Interval for one sample ttest: (95%) X t s crit X Formula (the same goes when working with zcritical values): mu = 44 +/ (2.306)(1.667) = 40.156 and 47.844 I am 95% confident that 40.156 and 47.844 contains the true population mean test score for the NYU college students. Now we ask: does the null hypothesis (0) fall within this interval No. So this affirms that we reject the null hypothesis. TWO Group TTests (Computation): Example: A researcher is interested in whether or not listening to classical music while studying affects a student’s performance on an exam. She creates 2 samples with 20 individuals in each, where Group 1 has listens to classical music while studying and Group 2 listens to rock, and compares the exam scores for each group. Group 1 Group 2 M=91.0 M=89.0 SS=5.88 SS = 4.88 Standard dev = 4 Standard dev = 3 6. Hypotheses: Null: M1 – M2 = 0 There is no difference in exam performance for students who listen to both classical and rock music. Alternative: M1 – M2 ≠ 0 There is a difference in exam performance for students who listen to both classical and rock music. 7. Calculating VARIANCES and STANDARD ERROR: (This will come in handy towards the end of the problem) Variance: We are given SS (sum of squares) and N. In order to find variance, we do: SS/df. [Note: with 2 sample ttests, df = (n1 + n2) – 2 ] So our variances for Groups 1 and 2 are 0.65 and 0.54, respectively. Standard Error: Plug these variances into the standard error formula to get 0.3450. 8. Decide on the appropriate test used to calculate t: Questions to ask: a. Are both sample sizes large (each sample size must be > 100) Yes – use large sample test for independent means (use z test!) No – Go to Question # 2 b. Are the sample sizes equal Yes – use pooled variances test for equal sample sizes No – Go to question #3 to check for Homogeneity of Variance [for there to be HoV, one variance has to be no more than twice as big as the other variance.] c. Can the population variances be assumed equal Yes – use pooled variances test No – use separate variances ttest NOTE: (mu1 – mu2) in the formulas will be replaced with 0 In this example, our sample sizes are small and equal use pooled variances test for equal sample sizes. After using this formula, we get a tcalculated score of 5.79. To find tcritical, we match up alpha (0.05) with the df (18) on table A.2. From the table, our tcritical is 2.101. 9. Compare tcalc with tcrit: If tcalc > tcrit, then we reject the null. If tcalc < tcrit then we fail to reject the null. Our tcalc (5.79) > tcrit (2.101), so we reject the null. Because we’re rejecting, we risk making a Type 1 Error. APA WRITE UP: Consistent with our predictions, the group that listened to classical music averaged higher in exam scores (M = 91.0, SD = 4) than did the group listening to rock music (M = 89.0, SD = 3), t(18) = 5.79, p < .05, twotailed). 10. Confidence Interval for 2 sample ttest: (95%) mu1 – mu2 = (91 89) +/ (2.101)(0.3450) = 1.2751 and 2.7248 I am 95% confident that 1.2751 and 2.7248 contains the population mean difference in exam scores between those who listened to classical music and those who listened to rock music. Now we ask: does the null hypothesis (0) fall within this interval No. So this affirms that we reject the null hypothesis. READING SPSS OUTPUT: Sig. = p level. So if sig. < alpha, then we reject the null! Levene’s test: if sig. > alpha, we assume there’s homogeneity of variance. But if sig. < alpha, we don’t assume homogeneity of variance POWER: Beta: (B) the probability of making a Type 2 error from an Alternative Hypothesis Distribution (which is the ttest distribution). Failing to reject when you should’ve rejected. Power: (1Beta) the probability that we correctly rejected the null hypothesis, and correctly found significant results. In other words, finding a difference when there really is a difference. We want bigger power! (0.08 and higher is good power) Delta – the expected tvalue. It is found on table A.4 using power and alpha. The first term depends on sample size and the second term is called effect size. Effect size – a measure of overlap between two population distributions. The difference between the 2 population means in terms of standard deviations [(mu1 – mu2) / standard dev.)] Small effect size: 0.2 Medium: 0.5 Large: 0.8 Relationship Sum Up: As beta decreases, then alpha increases, and power increases If power increases, then Type II error decreases As alpha increases, power increases (But…increasing alpha risks Type I error, so it’s easier to increase sample size instead). As sample size increases, power increases (statistical accuracy for rejecting the null) When you have a small t and a large N, effect size is small. But as N decreases, effect size increases (more room to overlap because there’s less room for variability). As sample size increases, delta increases, and power increases. Note: effect size and sample size affect power SEPARATELY. They don’t affect each other in the process of affecting power. 3 different Computation problems (looking for n, for effect size, and for power): 1. A researcher wants to compare GPAs between students who check some form of social media more than 20 times/day and students who check social media less than 20 times/day. If the researcher has already calculated the effect size (d) of 0.40, what n would she need to attain a power of 0.75, if alpha = 0.05, two tailed Two samples, so we will work with the 2 sample n formula. Power is 0.75 and alpha is 0.05, so go to table A.4 and use that to find delta. Delta should be 2.63. Plug delta and effect size into the formula to find n (n=87 participants). 2. If a researcher wants power to be 0.7 or more and n=10, what should the effect size be Power is 0.7 and alpha is 0.05, so go to table A.4 and find delta (2.48). Plug in delta and n into effect size formula. So d = 1.109. 3. Solve for power when given the following: n1=12 n2=12 M1=6.2 M2=5.4 Variance1=13.4 variance2=15.1 alpha = .01, 2tailed We are not given effect size (d), so instead, use the g formula: X 1 X 2 g sp The g formula has a pooled standard deviation in the denominator. Since we are given variance, we plug it into the pooled variance formula to find the pooled variance, and then take the square root to find the pooled standard deviation. Plug in the pooled standard deviation into g formula. Solve for g. (g=0.212) Use this g as your estimated d. Plug the g into the delta formula in place of d. Delta = 0.519. Go to table A.3. Align delta with alpha to find power. Power = 0.02 (extremely small)