x1 x2 x3 16 14 22 18 15 18 11 22 15 10 21 18 19 15 20 19 15 Test at a = 0.05 to determine if the population means are all the same. %3D 1. The null hypothesis is Ho: %3D Oµ1 = #2 = H3 = H4 = H5 O µi = H2 == #3 %3D 2. This is a right tailed test with: d.f.n = and d.f.d% = %3D %3D 3a. The STS (to 2 decimals) is F = 3b. The P-value (to 3 decimals) is: 4a. At a = 0.05, the decision is: O Reject Ho Do not reject Ho O Accept Ho Accept Ha

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4b. The conclusion is:
O There is sufficient evidence to conclude that not all means are equal
O There is insufficient evidence to conclude that not all means are equal
O There is sufficient evidence to conclude that all the means are equal
OThere is insSufficient evidence to conclude that all the means are equal
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Transcribed Image Text:4b. The conclusion is: O There is sufficient evidence to conclude that not all means are equal O There is insufficient evidence to conclude that not all means are equal O There is sufficient evidence to conclude that all the means are equal OThere is insSufficient evidence to conclude that all the means are equal Submit Question MacBook Air
Question 11
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x3
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Test at a = 0.05 to determine if the population means are all the same.
1. The null hypothesis is Ho:
O µi = µ2 = H3 = H4
Oµ1 = µ2 = H3 = H4 = µ5
O Hi = µ2 = H3
2. This is a right tailed test with:
d.f.n =
and d.f.d =
3a. The STS (to 2 decimals) is F =
3b. The P-value (to 3 decimals) is:
4a. At a = 0.05, the decision is:
O Reject Ho
O Do not reject Ho
Accept Ho
O Accept Ha
MacBook Air
Transcribed Image Text:Question 11 > x1 x2 x3 16 14 22 18 15 18 11 22 15 10 21 18 19 15 20 19 15 Test at a = 0.05 to determine if the population means are all the same. 1. The null hypothesis is Ho: O µi = µ2 = H3 = H4 Oµ1 = µ2 = H3 = H4 = µ5 O Hi = µ2 = H3 2. This is a right tailed test with: d.f.n = and d.f.d = 3a. The STS (to 2 decimals) is F = 3b. The P-value (to 3 decimals) is: 4a. At a = 0.05, the decision is: O Reject Ho O Do not reject Ho Accept Ho O Accept Ha MacBook Air
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