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The graph and Excel output below are for home runs distances (in feet) achieved by three baseball players. 550 500 450 400 35(1) We wish to know [Select] |(2) Ho: The overall average distances [Select (3) Ha: The overall average distances [Select] (4If we are making an error here, what type of error would it be? [Select

The graph and Excel output below are for home runs distances (in feet) achieved by three baseball players. 550 500 450 400 350 300 Player 2 Player 3 Player 1 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Player 1 10 4037 403.7 1290.23333 3072 438.857143 1853.47619 Player 2 7 Player 3 3802 422.444444 543.027778 ANOVA F crit Source of Variation SS MS P-value Between Groups 2598.35263 5196.70525 2.2071025 0.13278532 3.42213221 Within Groups 27077.1794 23 1177.26867 Total 32273.8846 25 Is it possible that the overall mean home run distances are about the same for these players? Report an appropriate hypothesis test with a 5% significance level. Write out the full procedure taught in this Home run distance (ft)
(1) We wish to know [Select] |(2) Ho: The overall average distances [Select (3) Ha: The overall average distances [Select] (4) (i) Select] (ii) [Select] [Select] are independent SRSS, and we (iii) We assume that the sample of assume that the population distribution of [Select] are approx Normal. df2= (5) F = [Select df1 Select] Select] Select P-value = [Select] then due to sampling variation alone, at (6) If the population average of all possible values of the F-statistics are expected to be most [Select] [Select ] [Select .This percentage is
If we are making an error here, what type of error would it be? [Select
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Answer #1

( Note - Some of the blank at not able to be solved without the options

These answer may not may the exact wordings of the options, hence please check which option is closed to the answer given)

1. We wish to know if overall mean home run distance are about the same for the players.

2. The overall average distance is the same
3. The overall average distance is not the same.

5

F = 2.207
df1 = 2
df2 = 23
Pvalue = 0.1327

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