Statistical Analysis Homework Solutions: T-Test and Confidence Intervals - Prof. Steven P., Assignments of Introduction to Sociology

Download Statistical Analysis Homework Solutions: T-Test and Confidence Intervals - Prof. Steven P. and more Assignments Introduction to Sociology in PDF only on Docsity! Socy 601 Answers to Homework #5 Question (23-58) Each (including the sub-question) is 10 points (8*10=80 points), Supplement questions (5*4=20 points) Chapter 6 23) (See Stata Printout below) # n df Confidence IntervaI 2-tailed p-value 1-tailed p-value t-score 23a) 5 4 95 0.05 0.025 2.776 23b) 15 14 95 0.05 0.025 2.145 23c) 25 24 95 0.05 0.025 2.064 23d) 26 25 95 0.05 0.025 2.06 23e) 26 25 99 0.01 0.005 2.787 25a) (See Stata Printout) Assumptions: 1. Random sampling methodology 2. Observations are independent 3. Number of days missed is or can be treated as an interval scale variable 4. Normal sampling distribution of reported days missed (n<30) Calculate standard error (Y(bar)): s / √(n) = 4 / √(20) = .8944 Calculate 95% confidence interval: Y(bar) +/- t.025 Y(bar) = 4 +/- (2.093)(.8944) = (2.1, 5.9) 25b) Does the normality assumption seem plausible for these data? Explain and discuss the implications of the term “robustness” for your analysis. No, the assumption of normality for the distribution of the reported days missed is not plausible for these data because it would imply that a significant number (15.87%) of students were absent for less than 0 days (which is an invalid statement). Generally, when n<30, we cannot assume a normal distribution of samples. However, the t-test is robust (i.e., performs well despite violated assumptions) to violations of normality if the population distribution is bell-curvish, so the results are meaningful. 26a) (See Stata Printout) This printout is a test of the null hypothesis that the treatment has no effect on infant heart rates. The sample size is 15 (Number of Cases, degrees of freedom (df) = 14). The sampled infants had a mean change in heart rate of 10.7 and a standard deviation (SD) of 17.7. The SE of Mean = SD/sqrt(n) = 17.7/sqrt(15) = 4.57. The sample mean (10.7) is 2.341 (t-value) SEs from the null mean of 0, which, for a sampling distribution of samples of 15, correlates to a p-value of .0346 (2-Tail Sig). In other words, 3.46% of samples with an n of 15 will have a sample mean that is 2.341 SEs from the null mean if the null mean is true. To find the 95% confidence interval, we use the t-value associated with a p of .05 (=2.145, see #23b) and multiply it by the SE = 4.570*2.145. We add and subtract this value from the sample mean to find the upper and lower bounds of the c.i. We are 95% confident that this interval contains the true population mean. 25b) Test whether the true mean is 0. To test whether the true mean is 0, we compare the probability that we would get a sample mean of 10.7 if the true mean were 0 against the risk we are willing to take of committing a Type 1 error. P(one-sided) = P(two-sided)/2 = .0346/2 = .0173. This value represents the probability that, if the null were true, we would get a t-score for samples of 15 that would be this large or larger. The two-tailed p-value is the probability that, if the null were true, we would get a t-score for samples of 15 that would be this far or farther from 0. 27) As sample size increases, a) the standard error decreases, and b) the t-score decreases, both of which result in a narrower confidence interval 33. A jury list contains the names of all individuals who may be called for jury duty. The proportion of the available jurors on the list who are women is .53. Test the hypothesis that the selections are random, if no woman is selected out of a sample of size 12. (See Stata Printout) If H0 is true, only .0116 % of samples (juries) would have no women. In other words, it is highly unlikely that the jury would have no women jury members. We reject the null hypothesis that this sample jury was truly drawn at random from a population that a list that was 53% female. 58. a. For each test, the probability equals .05 of falsely rejecting H0 and committing a Type I error. For 20 tests, the expected number of false rejection will be 20*.05 =1. That is, we expect about 1 researcher to incorrectly reject H0, but then this would be the one paper that would get published. This policy encourages the publication of papers suffering from Type I errors over other papers with “true” results. b. Of all the studies conducted, the one with the most extreme or unusual results is the one that gets substantial attention. That result is often the consequence of an unusual sample that, due to sampling error, is not representative of the population, with the sample mean far from the actual population mean. Further studies in later research would reveal that the true mean is not so extreme. 2 One-sample t test` ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 10 .55 .3162278 1 -.1653569 1.265357 ------------------------------------------------------------------------------ mean = mean(x) t = 1.7393 Ho: mean = 0 degrees of freedom = 9 Ha: mean < 0 Ha: mean != 0 Ha: mean > 0 Pr(T < t) = 0.9420 Pr(|T| > |t|) = 0.1160 Pr(T > t) = 0.0580 . F.) Briefly state a conclusion. We see that P=.116 for a two-tailed t-test, therefore we fail to reject the H0 that population has a neutral opinion about outcomes of preschool children of working mothers. 5 Stata Printout . log using hw5log, text replace (note: file C:\data\hw5log.log not found) --------------------------------------------------------------------------------------------------------------------- ----------------------- log: C:\data\hw5log.log log type: text opened on: 16 Oct 2008, 10:32:36 . . *23. . . di invttail(4,.025) 2.7764451 . di invttail(14,.025) 2.1447867 . di invttail(24,.025) 2.0638986 . di invttail(25,.025) 2.0595386 . di invttail(25,.005) 2.7874358 . . *25 . *Standard Error . di 4/sqrt(20) .89442719 . *Test Statistic . di invttail(19,.025) 2.0930241 . *Upper Bound . di 4+4/sqrt(20)*invttail(19,.025) 5.8720576 . *Lower Bound . di 4-4/sqrt(20)*invttail(19,.025) 6 2.1279424 . . *26 . ttesti 15 10.7 17.7 0 One-sample t test ------------------------------------------------------------------------------ | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval] ---------+-------------------------------------------------------------------- x | 15 10.7 4.57012 17.7 .8980667 20.50193 ------------------------------------------------------------------------------ mean = mean(x) t = 2.3413 Ho: mean = 0 degrees of freedom = 14 Ha: mean < 0 Ha: mean != 0 Ha: mean > 0 Pr(T < t) = 0.9827 Pr(|T| > |t|) = 0.0345 Pr(T > t) = 0.0173 . . *33 . bitesti 12 0 .53 N Observed k Expected k Assumed p Observed p ------------------------------------------------------------ 12 0 6.36 0.53000 0.00000 Pr(k >= 0) = 1.000000 (one-sided test) Pr(k <= 0) = 0.000116 (one-sided test) Pr(k <= 0) = 0.000116 (two-sided test) note: upper tail of two-sided p-value is empty . . log close log: C:\data\hw5log.log log type: text closed on: 16 Oct 2008, 10:32:36 ------------------------------------------------------------------------------------------------------------- 7