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# STAT 200 Week 6 Homework

LU

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Date Created: 11/06/15

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 0.01. 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 b c 4 6 7 3 7 8 2 8 5 1 4 7 4 6 9 2 4 2 a. Compute L for each subject using the contrast weights -1, 0, and 1. That is, compute (-1)(a) + (0)(b) + (1)(c) for each subject. L1 -1*4+0*6+1*7=3 L2 -1*3+0*7+1*8=5 L3 -1*2+0*8+1*5=3 L4 -1*1+0*4+1*7=6 L5 -1*4+0*6+1*9=5 L6 -1*2+0*4+1*2=0 b. Compute a one-sample t-test on this column (with the L values for each subject) you created. M=Sample Mean = (3+5+3+6+5+0)/6 = 3.667 Standard error of mean = Sm = 2.160/sqrt(6) = 0.8819 t=(M-mu)/Sm = 3.667/0.8819 = 4.158 Using calculator, we find out the probability of two tailed test to be 0.0088 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 first 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 significantly different at the .05 level? n1 = 12 , x1 = 75 , s1 = 10.954 n2 = 12 , x2 = 86 , s2 = 10 H0: u1 = u2 H1: u1 < u2 a. Pooled variance Sp^2 = 109.9951 Sp = 10.4879 .Standard error, SE = Sp * sqrt((1/n1)+(1/n2)) SE = 4.2816 .Test statistic, t = (x1-x2) / SE t = (75-86) / 4.28165 = -2.569 Critical value = -t(a,n1+n2-2) = - t(0.05, 22) Critical value = -1.717 Since t < critical value, hence reject H0 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 significant at the .05 level? Pooled variance Sp^2 = 109.9951 Sp = 10.4879 .Standard error, SE = Sp * sqrt((1/n1)+(1/n2)) SE = 6.0552 .Test statistic, t = (x1-x2) / SE t = (75-86) / 6.05516 = -1.817 .Critical value = -t(a,n1+n2-2) = - t(0.05, 10) Critical value = -1.812 .Since t < critical value, hence reject H0 4. Rank the following in terms of power. Population 1 N Population 2 Standard Power Mean Mean Deviation using 0.05 level of significan ce and two sided test A 29 20 43 12 1.00 B 34 15 40 6 0.97 C 105 24 50 27 1.00 D 170 2 120 10 1.00 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 first three has power of 1.00. .65. Previously, an organization reported that teenagers spent 4.5 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 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. Option D is correct 71. Previously, an organization reported that teenagers spent 4.5 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 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is: a. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher b. to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same (This option is correct) c. to conclude that the mean hours per week currently is 4.5, when in fact, it is higher d. to conclude that the mean hours per week currently is no higher than 4.5, 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 4.5 years, on average, to finish 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 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level? μ0=4.50 X−¿ 5.1 S =1.2∧n=49 x We need to check the sample mean is greater than 4.50. The t-test statistic is calculated as follows: 5.1−4.50 0.60 t= 1.2 = 0.1714=3.50 √ 49 The probability using calculator with test statistics of 3.50 and degree of freedom of 48 is: 0.0005 As, p value is less than 1% level of significance, so we reject the claim. 80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute 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. Nullhypot hesis: 1 −μ 20 Alternate Hypothesis:μ −μ =0 1 2 Sample Mean=X−1=32.375 ¯ SampleMean=X−2=30.625 s1=9.620∧s2=8.331 X1-bar minus X2-bar is the difference in the sample mean of girls relaxed time and jumping time. The distribution use is t-test with degree of freedom of 7. s1 s22 9.622 8.331 SE= √ n1+ n2 =√ 8 + 8 =4.50 Test Statistic=−1.51 Using calculator, the p value comes out to be 0.1755. 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 µ1 = 45 s1 = 14 n1 = 36 Powder Diet µ2 = 42 s2 = 12 n2 = 49 s = [s1^2 / n1 + s2^2 / n2]^0.5 = [14^2 / 36 + 12^2 / 49]^0.5 = 2.8954 We are testing H0: µ1=µ2 vs 1 : µ1>µ2 Under H 0 [(µ1 - µ2) – 0]/s Test Statistic = [(45 - 42) – 0] / 2.8954 = 1.0361 This is lesser than the critical value. So we have insufficient evidence to reject H0at 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 players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. 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= 2.25 μd Sample standard 3.775 deviation Distribution of the test is: Nullhypot hesis: μ =0 d Alternatehypot hesisd μ ≠0 Test statistics=25−0=1.19 3.775 4 √ The p value using calculator is 0.3196 Assuming 5% level of significance, p value is greater than level of significance, it means do not reject the null hypothesis and conclude that there is no difference in mean scores.

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