Question

The table available below shows the weights (in pounds) for a sample of vacuum cleaners. The weights are classified according to vacuum cleaner type. At a = 0.05, can you conclude that at least one mean vacuum cleaner weight is different from the others? ES: Click the icon to view the vacuum cleaner weight data. Let PBU. PALU, and Htc represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? O A. Ho: PBU = PBLU - PTC Weight of Vacuum Cleaners by Type -X H:HBUHBLU *HTC 23 26 24 21 20 25 OB. Ho: HBU = PBLU = HTC H: At least one of the means is different. Oc. Họ: HBU* HBLU+ HTC H: HBU = PBLU = HTC OD. Ho: At least one of the means is different. HEPBU = PBLU = HTC Bagged upright Bagless upright Top canister 23 26 26 23 24 22 25 24 23 27 28 22 Print Done Click to select your answer(s).

Answer 1

I am using Excel and the command is as follows

Data > Data Analysis > Anova : Single factor > select the data > ok

OUTPUT

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Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
BU	6	139	23.16667	5.366667
BLU	6	144	24	2.8
TC	6	149	24.83333	5.366667
ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	8.333333	2	4.166667	0.92	0.419
Within Groups	67.66667	15	4.511111
Total	76	17

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B)

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F = 0.92

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P-value = 0.419

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D)