A direct-mail company assembles and stores paper products (envelopes, letters, brochures, order cards, etc.) for its customers. The company estimates the total number of pieces received in a shipment by calculating the weight per piece and then weighing the entire shipment. The company is unsure whether the sample of pieces used to estimate the mean weight per piece should be drawn from a single carton or whether it is worth the extra time required to pull a few pieces from several cartons. To aid management in making a decision, eight brochures were removed from each of the five cartons of a typical shipment and weighed. The weights (in pounds) are displayed in the table.

| Carton 1 | Carton 2 | Carton 3 | Carton 4 | Carton 5 |
|----------|----------|----------|----------|----------|
| $.01851$ | $.01872$ | $.01869$ | $.01899$ | $.01882$ |
| $.01829$ | $.01861$ | $.01853$ | $.01917$ | $.01895$ |
| $.01844$ | $.01876$ | $.01876$ | $.01852$ | $.01884$ |
| $.01859$ | $.01886$ | $.01880$ | $.01904$ | $.01835$ |
| $.01854$ | $.01896$ | $.01880$ | $.01923$ | $.01889$ |
| $.01853$ | $.01879$ | $.01882$ | $.01905$ | $.01876$ |
| $.01844$ | $.01879$ | $.01862$ | $.01924$ | $.01891$ |
| $.01833$ | $.01879$ | $.01860$ | $.01893$ | $.01879$ |

**d**. Use Tukey's method to compare all pairs of means, with $\alpha=.05$ as the overall significance level.

Question

A direct-mail company assembles and stores paper products (envelopes, letters, brochures, order cards, etc.) for its customers. The company estimates the total number of pieces received in a shipment by calculating the weight per piece and then weighing the entire shipment. The company is unsure whether the sample of pieces used to estimate the mean weight per piece should be drawn from a single carton or whether it is worth the extra time required to pull a few pieces from several cartons. To aid management in making a decision, eight brochures were removed from each of the five cartons of a typical shipment and weighed. The weights (in pounds) are displayed in the table.

| Carton 1 | Carton 2 | Carton 3 | Carton 4 | Carton 5 |
|----------|----------|----------|----------|----------|
| $.01851$ | $.01872$ | $.01869$ | $.01899$ | $.01882$ |
| $.01829$ | $.01861$ | $.01853$ | $.01917$ | $.01895$ |
| $.01844$ | $.01876$ | $.01876$ | $.01852$ | $.01884$ |
| $.01859$ | $.01886$ | $.01880$ | $.01904$ | $.01835$ |
| $.01854$ | $.01896$ | $.01880$ | $.01923$ | $.01889$ |
| $.01853$ | $.01879$ | $.01882$ | $.01905$ | $.01876$ |
| $.01844$ | $.01879$ | $.01862$ | $.01924$ | $.01891$ |
| $.01833$ | $.01879$ | $.01860$ | $.01893$ | $.01879$ |

**d**. Use Tukey's method to compare all pairs of means, with $\alpha=.05$ as the overall significance level.

Quizlet Admin · Accepted Answer

Given:

$$\begin{aligned}
\alpha&=	ext{Significance level}=0.05
\ N&=	ext{Total sample size}=40
\ n_i&=	ext{Individual sample size}=8
\ k&=	ext{Number of treatment groups}=5
\&	extcolor{#4559AC}{	ext{Result previous part of this exercise:}}
\ MSE&=0
\end{aligned}$$

We need to compare all means using Tukey's method.

A pair of means are significantly different when they differ by more than $\omega=q_{\alpha}(k,v)\dfrac{s}{\sqrt{n_t}}$.

Baker's yeast:	Mean yield $(\%):$	$41.1$	$\overline{47.5}$	$48.6$	$50.3$
Brewer's	Temperature $\left({ }^{\circ} \mathrm{C}\right):$	54	45	48	51
yeast:	Mean yield $(\%):$	$39.4$	$47.3$	$49.2$	$49.6$
	Temperature $\left({ }^{\circ} \mathrm{C}\right):$	54	51	48	45

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Related questions

Carton 1	Carton 2	Carton 3	Carton 4	Carton 5
.01851	.01872	.01869	.01899	.01882
.01829	.01861	.01853	.01917	.01895
.01844	.01876	.01876	.01852	.01884
.01859	.01886	.01880	.01904	.01835
.01854	.01896	.01880	.01923	.01889
.01853	.01879	.01882	.01905	.01876
.01844	.01879	.01862	.01924	.01891
.01833	.01879	.01860	.01893	.01879