Question

Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 42 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 14 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 13,960; SSTR = 4,510.

(a)

Set up the ANOVA table for this problem. (Round your values for
MSE and *F* to two decimal places, and your *p*-value
to four decimal places.)

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

Treatments | ? | ? | ? | ? | ? |

Error | ? | ? | ? | ||

Total | ? | ? |

(b)

Use *α* = 0.05 to test for any significant difference in
the means for the three assembly methods.

State the null and alternative hypotheses.

*H*_{0}: *μ*_{1} ≠
*μ*_{2} ≠ *μ*_{3}

*H*_{a}: *μ*_{1} =
*μ*_{2} = *μ*_{3}

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

*H*_{a}: *μ*_{1} =
*μ*_{2} =
*μ*_{3}

*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}

*H*_{a}: *μ*_{1} ≠
*μ*_{2} ≠ *μ*_{3}

*H*_{0}: *μ*_{1} =
*μ*_{2} = *μ*_{3}

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

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

t stat =

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

*p*-value =

State your conclusion.

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

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

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

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

Answer #1

(a)

The ANOVA Table isset up for the problem as follows:

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

Treatments | 4510 | k - 1 = 3 - 1 = 2 | 4510/2 = 2255.00 | 2255.00/242.31 = 9.31 | 0.0005 |

Error | 13960 - 4510 = 9450 | N - k = 42 - 3 = 39 | 9450/39 = 242.31 | ||

Total | 13960 | N - 1 = 42 - 1 = 41 |

(b)

Correct option:

**H _{0}: μ_{1} = μ_{2} =
μ_{3}**

H_{a}: Not all the population means are equal.

(c)

F stat = 2255.00/242.31 = **9.31**

(d)

Degrees of Freedom for numerator = 2

Degrees of freedom for denominator = 39

By Technology, p - value = **0.0005**

(e)

Correct option:

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

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