# hw03_solution - Homework#3 14.3(a The detailed calculation...

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Unformatted text preview: Homework#3 14.3 (a) The detailed calculation is in the excel file. Source of variation Degree of freedom Sums of square Mean square Statistic F Treatments 3 737.8868 245.9623 24.5962 Error 49 490 10 Total 52 1227.8868 (b) The detailed calculation is in the excel file. Source of variation Degree of freedom Sums of square Mean square Statistic F Treatments 3 737.8868 245.9623 24.5962 Error 49 490 10 Total 52 1227.8868 (c) Statistic F keeps unchanged while increasing the sample means by 100. 14.7 Defineμ1: average amount of spam email that professors received per day μ2: average amount of spam email that administrator received per day μ3: average amount of spam email that students received per day   2 1 :   H 3  : 1 H At least two means differ. Assume the three populations follow ND and we’ve known that the sample is drawn randomly from independent populations. Now we use Bartlett’s test to test for the equality of variances of the three populations (at 5% significance level):   2 2 1 2 :   H 3 2  : 1 H At least two variances differ. As p-value = 0.5937 > 0.05, we fail to reject the null hypothesis. That is, we could conclude with 97.5% confidence that there is insufficient evidence to say the variances of spam mail received by professor ,administrator and students differ. So now we can use one-way ANOVA to test the equality of the three population means.   2 1 :   H 3  : 1 H At least two means differ. [Using Excel Data Analysis tool] P value(0.44) > alpha(0.025), not reject H0. There is not enough evidence to conclude that the differing university communities differ in the amount of spam they receive in their emails. 14.9 (a) Let 1  , 2  , 3  and 4  be the average grades in the first year at the university of students from school A, B, C and D Assume that the four populations all follow ND and are with the same variances. We’ve known that the samples are drawn randomly from independent populations. Thus we could use one-way ANOVA to test the equality of the average grades of students from the four different schools. (Given that the significant level = 5%)   2 1 :   H 3  = 4  : 1 H At least two means differ. Rejection rule: if p-value < 0.05 (or statistic F > 76 . 2 77 , 3 , 05 .  F ), we could reject the null hypothesis. [Using Excel Data Analysis tool] 單因子變異數分析 摘要 組 個數 總和 平均 變異數 School A 20 1376.5 68.825 52.27987 School B 26 1692 65.07692 37.38265 School C 16 992.2 62.0125 63.45583 School D 19 1228.2 64.64211 56.88035 ANOVA 變源 SS 自由度 MS F P- 值 臨界值 組間 430.9506 3 143.6502 2.833578 0.043693 2.723343 組內 3903.567 77 50.69568 總和 4334.518 80 [Using simple excel function only] Let j X be the average grades in the first year at the university of students from school j (j = 1: A, j = 2: B, j = 3: C and j = 4: D) and X be the grand mean of the total sample data....
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## This note was uploaded on 01/06/2012 for the course MANAGEMENT 000 taught by Professor 游啟璋 during the Fall '11 term at National Taiwan University.

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hw03_solution - Homework#3 14.3(a The detailed calculation...

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