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TABLE 11-5 A Hotel Chain Has Identically Small Sized Resorts in 5

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TABLE 11-5
A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. Analysis of Variance TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. * or p < 0.005,tabular value
-Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is


A) H0 : μA = μB = μC = μD.
B) H0 : MA = MB = MC = MD.
C) H0 : TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. = TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. = TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. = TABLE 11-5 A hotel chain has identically small sized resorts in 5 locations in different small islands.The data that follow resulted from analyzing the hotel occupancies on randomly selected days in the 5 locations. Analysis of Variance * or p < 0.005,tabular value -Referring to Table 11-5,the null hypothesis for Levene's test for homogeneity of variances is A) H<sub>0</sub> : μ<sub>A</sub> = μ<sub>B</sub> = μ<sub>C </sub>= μ<sub>D</sub>. B) H<sub>0</sub> : M<sub>A</sub> = M<sub>B </sub>= M<sub>C </sub>= M<sub>D</sub>. C) H<sub>0 </sub>: = = = . D) H<sub>0 </sub>: π<sub>A</sub> = π<sub>B </sub>= π<sub>C</sub> = π<sub>D</sub>. .
D) H0 : πA = πB = πC = πD.

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