12.6 Review Questions
- The data on monthly rents, in dollars, for independent random samples of newly completed apartments in the four U.S. regions are presented in the following table.
Northeast Midwest South West 1005 870 891 1025 898 748 630 1012 948 699 861 1090 1181 814 1036 926 1244 721 1269 606 Given the ANOVA table, test at the 5% significance level whether a difference exists in the mean rent of newly completed apartments in the four U.S. regions.
Source df SS MS = [latex]\frac{SS}{df}[/latex] F-statistic P-value Region 3 SSTR = 400513 MSTR = 133504 [latex]F_o[/latex] = 7.541 0.0023 Error 16 SSE = 283265 MSE = 17704 Total 19 SST = 683778 - The following table gives the salaries (in thousand dollars) for computer science (CS) majors obtaining a bachelor’s degree, a master’s degree, or a Ph.D.
Bachelor Master PhD 50.8 65.8 73.3 59.4 57.5 65.7 55.9 66.9 71.7 45.1 62.8 72.5 54.1 68.5 73 50.7 69.3 67.2 46.8 61.5 67.5 - Fill in missing entries of the following ANOVA table.
Source df SS MS = [latex]\frac{SS}{df}[/latex] F-statistic P-value Group k-1 = ? SSTR = ? [latex]\text{MSTR} = \frac{SSTR}{k-1} = 616.9[/latex] [latex]F_o = \frac{MSTR}{MSE} = 34.62[/latex] <0.0001 Error n–k = ? SSE = ? [latex]\text{MSE} = \frac{SSE}{n-k} = 17.8[/latex] Total n-1 = 20 SST = 1554.5 - Test at the 1% significance level whether a difference exists in mean salary for computer science (CS) majors obtaining a bachelor’s degree, a master’s degree, or a Ph.D.
- Fill in missing entries of the following ANOVA table.
Show/Hide Answer
- From the first table, we have [latex]k=4, n_1=5, n_2=6, n_3=4, n_4=5 \longrightarrow n=n_1+n_2+n_3+n_4=20[/latex].
We assume the assumptions of the one-way ANOVA F test are satisfied. Steps of one-way ANOVA F test:
Step 1: Hypotheses.
[latex]H_0:[/latex] all means are equal, [latex]\mu_1=\mu_2=\mu_3=\mu_4[/latex]versus [latex]H_a:[/latex] not all means are equal, i.e., at least one pair of means are different.
Step 2: Significance level [latex]\alpha=0.05.[/latex]
Step 3: Test statistic [latex]F_o[/latex]=7.541 with degrees of freedom [latex]df_{TR}=k-1=4-1=3, df_E=n-k=20-4=16.[/latex]
Step 4: P-value=0.0023.
Step 5: Decision. Reject [latex]H_0[/latex] since P-value=0.0023 <0.01 [latex](\alpha)[/latex].
Step 6: Conclusion. At the 5% significance level, we have sufficient evidence that a difference exists in the mean rent of newly completed apartments in the four U.S. regions. -
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Source df SS MS=[latex]\frac{SS}{df}[/latex] F-statistic P-value Group k-1=? SSTR=? MSTR=[latex]\frac{SSTR}{k-1}=616.9[/latex] [latex]F_o=\frac{MSTR}{MSE}=34.62[/latex] <0.0001 Error n-k=? SSE=? MSE=[latex]\frac{SSE}{n-k}=17.8[/latex] Total n-1=20 SST=1554.5 From the first table, we have [latex]k=3, n_1=n_2=n_3=7 \Longrightarrow n=n_1+n_2+n_3=21[/latex].
From the second table, we have [latex]df_{T}=n-1=19, SST=1554.5, MSTR=616.0, MSE=17.8[/latex].
(1) [latex]df_{TR}=k-1=3-1=2.[/latex](2) [latex]df_E=n-k=21-3=18[/latex] or [latex]df_E=df_T-df_{TR}=19-2=17.[/latex]
(3) [latex]SSTR=MSTR\times df_{TR}=616.9\times 2=1233.8.[/latex]
(4) [latex]SSE=MSE \times df_E=17.8 \times 18=320.4[/latex] or [latex]SSE=SST-SSTR=1554.5-1233.8=320.7.[/latex] The difference is due to rounding.
- We assume the assumptions of the one-way ANOVA F test are satisfied.
Steps of one-way ANOVA F test:
Step 1: hypotheses.[latex]H_0:[/latex] all means are equal, [latex]\mu_1=\mu_2=\mu_3[/latex] versus [latex]H_a:[/latex] not all means are equal, i.e., at least one pair of means are different.
Step 2: Significance level [latex]\alpha=0.01.[/latex]
Step 3: Test statistic [latex]F_o=34.62[/latex] with degrees of freedom [latex]df_{TR}=k-1=3-1=2, df_E=n-k=21-3=18.[/latex]
Step 4: P-value <0.0001.
Step 5: Decision. Reject [latex]H_0[/latex] since P-value<0.0001 <0.01 [latex](\alpha)[/latex].
Step 6: Conclusion. At the 1% significance level, we have sufficient evidence that a difference exists in mean salary for computer science (CS) majors obtaining a bachelor’s degree, a master’s degree or a Ph.D.
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