Question

The following data were obtained for a randomized block design
involving five treatments and three blocks: SST = 510, SSTR = 370,
SSBL = 95. Set up the ANOVA table. (Round your value for *F*
to two decimal places, and your *p*-value to three decimal
places.)

Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F |
p-value |
---|---|---|---|---|---|

Treatments | |||||

Blocks | |||||

Error | |||||

Total |

Test for any significant differences. Use *α* = 0.05.

State the null and alternative hypotheses.

*H*_{0}: Not all the population means are
equal.

*H*_{a}: *μ*_{1} =
*μ*_{2} = *μ*_{3} =
*μ*_{4} =
*μ*_{5}*H*_{0}:
*μ*_{1} = *μ*_{2} =
*μ*_{3} = *μ*_{4} =
*μ*_{5}

*H*_{a}: *μ*_{1} ≠
*μ*_{2} ≠ *μ*_{3} ≠
*μ*_{4} ≠
*μ*_{5} *H*_{0}:
At least two of the population means are equal.

*H*_{a}: At least two of the population means are
different.*H*_{0}: *μ*_{1} ≠
*μ*_{2} ≠ *μ*_{3} ≠
*μ*_{4} ≠ *μ*_{5}

*H*_{a}: *μ*_{1} =
*μ*_{2} = *μ*_{3} =
*μ*_{4} =
*μ*_{5}*H*_{0}:
*μ*_{1} = *μ*_{2} =
*μ*_{3} = *μ*_{4} =
*μ*_{5}

*H*_{a}: Not all the population means are equal.

Find the value of the test statistic. (Round your answer to two decimal places.)

Find the *p*-value. (Round your answer to three decimal
places.)

*p*-value =

State your conclusion.

Do not reject *H*_{0}. There is sufficient
evidence to conclude that the means of the treatments are not all
equal.

Reject *H*_{0}. There is not sufficient evidence
to conclude that the means of the treatments are not all
equal.

Do not reject *H*_{0}. There is not sufficient
evidence to conclude that the means of the treatments are not all
equal.

Reject *H*_{0}. There is sufficient evidence to
conclude that the means of the treatments are not all equal.

Answer #1

Solution;-

Source of Variation |
Sum of squares |
Degree of Freedom |
Mean Square |
F | p-value |

Treatments | 370 | 4 | 92.5 | 16.444 | 0.001 |

Blocks | 95 | 2 | 47.5 | 8.4444 | 0.011 |

Error | 45 | 8 | 5.625 | ||

Total | 510 | 14 |

**State the hypotheses.** The first step is to
state the null hypothesis and an alternative hypothesis.

*H*_{a}: *μ*_{1} =
*μ*_{2} = *μ*_{3} =
*μ*_{4} = *μ*_{5}

*H*_{a}: At least two of the population means are
different.

**Formulate an analysis plan**. For this analysis,
the significance level is 0.05.

**Analyze sample data**.

F statistics is given by:-

**F = 16.444**

**The P-value = 0.001**

**Interpret results. Since the P-value (0.001) is less
than the significance level (0.05), we have to reject the null
hypothesis.**

**Conclusion:-**

**Reject H_{0}. There is sufficient
evidence to conclude that the means of the treatments are not all
equal.**

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