/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Three different assembly methods... [FREE SOLUTION] | 91Ó°ÊÓ

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Three different assembly methods have been proposed for a new product. A completely randomized experimental design was chosen to determine which assembly method results in the greatest number of parts produced per hour, and 30 workers were randomly selected and assigned to use one of the proposed methods. The number of units produced by each worker follows. $$\begin{array}{rrr} & \text { Method } & \\ \mathbf{A} & \mathbf{B} & \mathbf{C} \\ 97 & 93 & 99 \\ 73 & 100 & 94 \\ 93 & 93 & 87 \\ 100 & 55 & 66 \\ 73 & 77 & 59 \\ 91 & 91 & 75 \\ 100 & 85 & 84 \\ 86 & 73 & 72 \\ 92 & 90 & 88 \\ 95 & 83 & 86 \end{array}$$ Use these data and test to see whether the mean number of parts produced is the same with each method. Use \(\alpha=.05\)

Short Answer

Expert verified
The mean number of parts produced is not significantly different among the methods at the 0.05 significance level.

Step by step solution

01

State the Hypotheses

To determine if there are differences in the mean number of parts produced among the three assembly methods, we set up our hypotheses as follows: - Null Hypothesis ( H_0 ): The mean number of parts produced is the same for all methods (Method A = Method B = Method C). - Alternative Hypothesis ( H_a ): At least one method has a different mean.
02

Calculate the Group Means and Overall Mean

First, compute the mean number of units produced for each method:\[\bar{X}_A = \frac{97 + 73 + 93 + 100 + 73 + 91 + 100 + 86 + 92 + 95}{10} = 90\]\[\bar{X}_B = \frac{93 + 100 + 93 + 55 + 77 + 91 + 85 + 73 + 90 + 83}{10} = 84\]\[\bar{X}_C = \frac{99 + 94 + 87 + 66 + 59 + 75 + 84 + 72 + 88 + 86}{10} = 81\]Combine these to find the overall mean:\[\bar{X}_{\text{overall}} = \frac{(10 \times 90) + (10 \times 84) + (10 \times 81)}{30} = 85\]
03

Compute the Sum of Squares

Now calculate the Sum of Squares for Treatment (SST), Error (SSE), and Total (SSTotal):- SST measures the variation between the groups:\[SST = 10[(90-85)^2 + (84-85)^2 + (81-85)^2] = 10[25 + 1 + 16] = 420\]- To find SSE, compute the sum of squared deviations within each group:For Method A:\[\sum (X_{Ai} - \bar{X}_A)^2 = (97-90)^2 + (73-90)^2 + ... + (95-90)^2 = 790\]For Method B:\[\sum (X_{Bi} - \bar{X}_B)^2 = 780\]For Method C:\[\sum (X_{Ci} - \bar{X}_C)^2 = 586\]\[SSE = 790 + 780 + 586 = 2156\]- Total Variation:\[SSTotal = SST + SSE = 420 + 2156 = 2576\]
04

Perform ANOVA Test

Now we need to calculate the test statistic using ANOVA:- Calculate Mean Squares:\[MS_{\text{Treatment}} = \frac{SST}{df_{\text{Trt}}} = \frac{420}{2} = 210\]\[MS_{\text{Error}} = \frac{SSE}{df_{\text{Error}}} = \frac{2156}{27} \approx 79.85\]- Compute F-statistic:\[F = \frac{MS_{\text{Treatment}}}{MS_{\text{Error}}} = \frac{210}{79.85} \approx 2.63\]
05

Make a Decision

Compare the calculated F-statistic with the critical value from the F-distribution table with (dft = 2 dferror = 27 ) at alpha = 0.05 . The critical value is approximately 3.35. Since 2.63 is less than 3.35, we do not reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Experimental Design
When conducting a study to compare different methods or treatments, an effective plan must be in place to ensure valid results. This plan is known as an **experimental design**. For the problem at hand, a completely randomized design was chosen. This means that the 30 workers were randomly assigned to one of the three assembly methods (A, B, or C). The randomness helps to prevent biases and distribute variability equally among groups.

A well-structured experimental design allows researchers to attribute differences in outcomes to the methods being tested, rather than to external variables or biases. This is essential in determining the effectiveness of the different assembly methods in terms of productivity per hour.

Choosing the right design is crucial for reliability. In this exercise, using a random assignment helps ensure that any observed differences in the output are likely due to the methods themselves rather than unforeseen confounding variables.
Null Hypothesis
In statistical testing, we start by stating a **null hypothesis (Hâ‚€)**, which usually represents a status quo or no effect scenario. For ANOVA, this means hypothesizing that all group means are equal. In the exercise, the null hypothesis is that the mean number of parts produced per hour is the same for all three methods: A, B, and C (Method A = Method B = Method C).

The null hypothesis serves as a point of comparison to determine if the evidence is strong enough to suggest a statistically significant difference between groups. If the null hypothesis is rejected, it indicates that at least one group mean is different from the others, showing potential for a more effective assembly method.

Essentially, stating the null hypothesis helps in setting the framework for analysis and decision-making regarding whether the differences in productivity really exist.
Sum of Squares
The **Sum of Squares (SS)** is a crucial calculation in ANOVA that helps to quantify the variability within and between groups. It serves to determine whether there are significant differences in group means and involves the following:
  • **SST (Sum of Squares for Treatment):** Measures the variation between group means. It reflects how different each group's mean is from the overall mean.
  • **SSE (Sum of Squares for Error):** Accounts for variability within each group. This depicts how each individual's score differs from their group mean.
  • **SSTotal (Total Sum of Squares):** Represents the overall variability, combining both SST and SSE.
In our exercise, these calculations revealed SST = 420, SSE = 2156, and SSTotal = 2576. Understanding these concepts helps us to break down where variance is occurring within the data.
F-statistic
The **F-statistic** is a key outcome from an ANOVA test that helps us determine whether the differences between group means are statistically significant. It's calculated by dividing the Mean Square for Treatment (MST) by the Mean Square for Error (MSE).
  • **MST (Mean Square Treatment):** Obtained by dividing the SST by its degrees of freedom (df extsubscript{Trt}).
  • **MSE (Mean Square Error):** Found by dividing SSE by its degrees of freedom (df extsubscript{Error}).
The F-statistic helps make a decision on the null hypothesis. In our example: - F = MST/MSE = 210/79.85 ≈ 2.63 This ratio compares the variance between groups to the variance within groups. A larger F-statistic indicates a greater likelihood that the observed differences between groups are not due to random chance. However, in this scenario, our calculated F-statistic (2.63) was less than the critical value (3.35), leading to a decision not to reject the null hypothesis, suggesting no significant difference between the assembly methods.

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Most popular questions from this chapter

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