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## STAT 200 Week 6 Homework

by: kimwood Notetaker

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# STAT 200 Week 6 Homework

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STAT 200 Week 6 Homework
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This 0 page Study Guide was uploaded by kimwood Notetaker on Friday November 13, 2015. The Study Guide belongs to a course at a university taught by a professor in Fall. Since its upload, it has received 19 views.

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Date Created: 11/13/15
18 You choose an alpha level of 01 and then analyze your data a What is the probability that you will make a Type I error given that the null hypothesis is true The probability of type I error is actually alpha given that the null hypothesis is true so it is 001 b What is the probability that you will make a Type I error given that the null hypothesis is false When null hypothesis is false it is impossible to make a type I error It means probability that you will make a type I error given that the null hypothesis is false is zero 7 Below are data showing the results of six subjects on a memory test The three scores per subject are their scores on three trials a b and c of a memory task Are the subjects get ting better each trial Test the linear effect of trial for the data a Compute L for each subject using the contrast weights 1 O and 1 That is compute 1a Ob 1c for each subject 14O6173 13O7185 1208153 11O4176 1406195 1204120 b Compute a onesample ttest on this column with the L values for each subject you created MSample Mean 3536506 3667 Standard error of mean Sm 2160sqrt6 08819 tMmuSm 366708819 4158 Using calculator we nd out the probability of two tailed test to be 00088 13 You are conducting a study to see if students do better when they study all at once or in intervals One group of 12 participants took a test after studying for one hour continuously The other group of 12 participants took a test after studying for three twenty minute sessions The rst group had a mean score of 75 and a variance of 120 The second group had a mean score of 86 and a variance of 100 a What is the calculated t value Are the mean test scores of these two groups signi cantly different at the 05 level n1 12X1 7551 10954 n212x2865210 H0 u1 u2 H1 u1 lt u2 a Pooled variance Spquot2 1099951 3 I 51101 1139 5 Wa 11 F m Hg 2 Sp 104879 Standard error SE Sp sqrt1n11n2 SE 42816 Test statistic t x1X2 SE t 7586 428165 2569 Critical value tan1n22 t005 22 Critical value 1717 Since t lt critical value hence reject HO So we conclude that students do better when they study all at once or in intervals b What would the t value be if there were only 6 participants in each group Would the scores be signi cant at the 05 level Pooled variance Spquot2 1099951 Sp 104879 Standard error SE Sp sqrt1n11n2 SE 60552 Test statistic t x1X2 SE t 7586 605516 1817 Critical value tan1n22 t005 10 Critical value 1812 Since t lt critical value hence reject HO 4 Rank the following in terms of power Population 1 N Population 2 Standard Power Mean Mean Deviation using 005 level of signi can ce and two sided test A 29 20 43 12 100 B 34 15 40 6 097 C 105 24 50 27 100 D 170 2 120 10 100 The power of a test is the probability of correctly rejecting null hypothesis when it is false The power ranking based on above calculations is A C D B with rst three has power of 100 65 Previously an organization reported that teenagers spent 45 hours per week on average on the phone The organization thinks that currently the mean is higher Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone The sample mean was 475 hours with a sample standard deviation of 20 Conduct a hypothesis test The 1111 and alternative lflypetheeee are 31 HQ 1 151 Hg 1 e 45 b He p 3 45 Hg u i 45 C Hi3 p 45 Hg u e 45 d Hi p 45 Hg p e 45 Option D is correct 71 Previously an organization reported that teenagers spent 45 hours per week on average on the phone The organization thinks that currently the mean is higher Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone The sample mean was 475 hours with a sample standard deviation of 20 Conduct a hypothesis test the Type I error is a to conclude that the current mean hours per week is higher than 45 when in fact it is higher b to conclude that the current mean hours per week is higher than 45 when in fact it is the same This option is correct c to conclude that the mean hours per week currently is 45 when in fact it is higher d to conclude that the mean hours per week currently is no higher than 45 when in fact it is not higher 77 An article in the San Jose Mercury News stated that students in the California state university system take 45 years on average to nish their undergraduate degrees Suppose you believe that the mean time is longer You conduct a survey of 49 students and obtain a sample mean of 51 with a sample standard deviation of 12 Do the data support your claim at the 1 level 02450 X Z51 Sx12n49 We need to check the sample mean is greater than 450 The ttest statistic is calculated as follows t51 450 060 T 12 01714 g 2350 The probability using calculator with test statistics of 350 and degree of freedom of 48 is 00005 As p value is less than 1 level of signi cance so we reject the claim 80 At Rachel39s 11th birthday party eight girls were timed to see how long in seconds they could hold their breath in a relaxed position After a twominute rest they timed themselves while jumping The girls thought that the mean difference between their jumping and relaxed times would be zero Test their hypothesis eleeted time secede Juming time seentie ze 21 4 4H 3e 2 22 21 elexed time eeende Juming time see 23 25 43 3 Null hypot h esis ul uz i 0 Alternate Hypot h esis ul x12 0 Sample Mean 2 X 1 232375 Sample Mean 2 X 2 30625 S 19620As28331 Xlbar minus X2bar is the difference in the sample mean of girls relaxed time and jumping time The distribution use is ttest with degree of freedom of 7 512 522 9622 83312 ngj 2450 n1 n2 8 8 Test Statistic 151 Using calculator the p value comes out to be 01755 This is the probability that there is difference between the relaxed time and jumping time The test statistic and critical value results indicate that there is a difference between the relaxed time and jumping time So do not reject the null hypothesis 91 A powder diet is tested on 49 people and a liquid diet is tested on 36 different people Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds Liquid Diet u1 45 1 14 n1 36 Powder Diet u242 5212 n249 s 51quot2 n1 52quot2 n2quotO5 14quot2 36 12quot2 49 O5 28954 We are testing Ho u1u2 vs H1 u1gtu2 Under H0 lll H2 Os Test Statistic 45 42 O 28954 10361 This is lesser than the critical value So we have insufficient evidence to reject H0 at the 5 level Thus liquid diet is not more effective than powder diet so do not reject null hypothesis 120 A golf instructor is interested in determining if her new technique for improving players39 golf scores is effective She takes four new students She records their 18hole scores before learning the technique and then after having taken her class She conducts a hypothesis test The data are as follows lPlayer ill Player 2 Player 3 Player 4 Mean EC l39E befere Glass 83 TB 93 Mean ecpre after elaee 80 SD E 86 The given test is a matched pair test Difference in means Player 1 3 Player 2 2 Player 3 7 Player 4 1 Mean of difference 225 Ud Sample standard 3775 deviation Distribution of the test is Null hypot h esis ud 0 Alternate h ypot h esis ud 0 225 O Test statistics 2 119 The p value using calculator is 03196 Assuming 5 level of signi cance p value is greater than level of signi cance it means do not reject the null hypothesis and conclude that there is no difference in mean scores

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