Practice Problems free essay sample

Evolutionary theories often emphasize that humans have adapted to their physical environment. One such theory hypothesizes that people should spontaneously follow a 24-hour cycle of sleeping and waking—even if they are not exposed to the usual pattern of sunlight. To test this notion, eight paid volunteers were placed (individually) in a room in which there was no light from the outside and no clocks or other indications of time. They could turn the lights on and off as they wished. After a month in the room, each individual tended to develop a steady cycle. Their cycles at the end of the study were as follows: 25, 27, 25, 23, 24, 25, 26, and 25. Using the . 05 level of significance, what should we conclude about the theory that 24 hours is the natural cycle? (That is, does the average cycle length under these conditions differ significantly from 24 hours? ) (a) Use the steps of hypothesis testing. (b) Sketch the distributions involved, (c) Explain your answer to someone who has never taken a course in statistics. We will write a custom essay sample on Practice Problems or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Answer Null hypothesis H0: = 24 hours Alternative hypothesis: H1: ? 24 hours Df = 7 Critical t value = Â ±2. 36 Sample Mean = 25 Standard deviation = 1. 195 The test statistic used is P-value = 0. 049867231 Since calculated p-value 0. 049867231 is slightly less than 0. 05(significance level) therefore we reject the null hypothesis. Because there is enough evidence to support the claim that the average cycle length under experimental conditions is significantly different from 24 hours Ch. 8, Practice Problem: 18 Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the . 05 level, what should the experimenter conclude? (a) Use the steps of hypothesis testing, (b) sketch the distributions involved, and (c) explain your answer to someone who is familiar with the t test for a single sample but not with the t test for independent means. Answer Null hypothesis: Â µ1 = Â µ2 Alternate hypothesis: Â µ1 ? Â µ2 Pooled standard deviation = sqrt((19*3^2+29*5^2)/(20+30-2)) = 4. 32 t-statistic = (38 – 30)/(4. 32*sqrt(1/20+1/30)) = 2. 4056 degree of freedom = 20+30-2 = 48 t-critical = 2. 0106 As t-statistic is greater than critical t, we have rejected the null hypothesis. The scores for two groups are not same. The group that was been experiment on had higher scores. There is enough evidence to support that there is significant difference in mean knowledge of two groups. Ch. 9, Practice Problem: 17 Do students at various universities differ in how sociable they are? Twenty-five students were randomly selected from each of three universities in a region and were asked to report on the amount of time they spent socializing each day with other students. The result for University X was a mean of 5 hours and an estimated population variance of 2 hours; for University Y, M = 4, S2 = 1. 5; and for University Z, M = 6, S2 = 2. 5. What should you conclude? Use the . 05 level. (a) Use the steps of hypothesis testing, (b) figure the effect size for the study; and (c) explain your answers to parts (a) and (b) to someone who has never had a course in statistics. Answer Null hypothesis: H0: There is no significant difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges. Alternative hypothesis: H1: There is significant difference in the mean amount of time spent for socializing each day with other students by student from 3 different colleges. The test statistic used is F test (ANOVA). Conclusion: Reject the null hypothesis, the value of test statistic is greater than the critical value of F. The sample provides enough evidence to support the claim that there is significant difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges. Effect Size We calculate R^2 = 50/350 = 0. 143. Thats a measure of effect size. ANOVA test is used test whether there is difference in the mean amount of time spent for socializing each day with other students by student from 3 colleges. The Value of test statistic is greater than the critical value indicate that there is significant difference in the mean values. The effect size for the study is 0. 25. Thus 25% of the total variance in the amount of time spent for socializing each day with other students is attributed to the college. C. This means that there is good reason to think that not all of the colleges are the same. Furthermore, we can explain around 14% of the variability in time spent socializing. Ch. 11, Practice Problems: 11 Make up a scatter diagram with 10 dots for each of the following situations: 1. perfect positive linear correlation, 2. large but not perfect positive linear correlation, 3. small positive linear correlation, 4. large but not perfect negative linear correlation, 5. no correlation, 6. clear curvilinear correlation. For problems 12 to 14, do the following: 1. Make a scatter diagram of the scores; 2. describe in words the general pattern of correlation, if any; 3. figure the correlation coefficient; 4. figure whether the correlation is statistically significant (use the . 05 significance level, two-tailed); 5. explain the logic of what you have done, writing as if you are speaking to someone who has never heard of correlation (but who does understand the mean, deviation scores, and hypothesis testing); and 6. give three logically possible directions of causality, indicating for each direction whether it is a reasonable explanation for the correlation in light of the variables involved (and why). Answer A. Perfect positive linear correlation B. large but not perfect positive linear correlation C. small positive linear correlation D. large but not perfect negative linear correlation E. no correlation F. clear curvilinear correlation Ch. 11, Practice Problems: 12 Four research participants take a test of manual dexterity (high scores mean better dexterity) and an anxiety test (high scores mean more anxiety). The scores are as follows. Answer A. B. We see a negative relationship between the two variables. The correlation between dexterity and anxiety is negative. Higher values of dexterity tend to be associated with low values of Anxiety (and vice versa). C. Correlation coefficient r = -0. 9037 correlation coefficient from excel using the =correl () function. D. Ho: Correlation coefficient is greater than or equal to zero Ha: Correlation coefficient is less than zero Test statistic t = -2. 98481 P-value = 0. 048152 Since p-value is less than 5% we reject Ho and conclude that correlation coefficient is negative. E. Since the correlation was zero, what we have found is that seeing the data we have observed is highly unlikely. Concluding the variables are related in a significant way. If there is increase in dexterity there is decrease in anxiety. F. It might be the case that high dexterity is a cause of low anxiety. That is, the more skilled one is with ones hands, and the less likely they are to be anxious. On the other hand, it might be the case that high anxiety causes dexterity skills to be low. That is, a nervous person might have shaking hands, resulting in poor dexterity scores. The third possibility is that both variables are caused by a third unknown variable. That is, it might be the case that the tests were given under different time limits. Hence the more time you are given the less anxious you are to be and the more dexterous you are able to be because you have more time.