91Ó°ÊÓ

In an experiment to assess the effect of the angle of pull on the force required to cause separation in electrical connectors, four different angles (factor \(A\) ) were used and each of a sample of five connectors (factor \(B\) ) was pulled once at each angle ("A Mixed Model Factorial Experiment in Testing Electrical Connectors," Indust. Qual. Control, 1960: 12-16). The data appears in the accompanying table. $$ \begin{array}{cc|ccccc} & & {\boldsymbol{B}} \\ & & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline & \mathbf{0}^{\circ} & 45.3 & 42.2 & 39.6 & 36.8 & 45.8 \\ \mathbf{A} & \mathbf{2}^{\circ} & 44.1 & 44.1 & 38.4 & 38.0 & 47.2 \\ & \mathbf{4}^{\circ} & 42.7 & 42.7 & 42.6 & 42.2 & 48.9 \\ & \mathbf{6}^{\circ} & 43.5 & 45.8 & 47.9 & 37.9 & 56.4 \\ \hline \end{array} $$ Does the data suggest that true average separation force is affected by the angle of pull? State and test the appropriate hypotheses at level \(.01\) by first constructing an ANOVA table (SST \(=396.13, \mathrm{SSA}=58.16\), and \(\mathrm{SSB}=246.97\) ).

Step by step solution

01

Setup Hypotheses

To determine if the angle affects the separation force, we set up the null hypothesis \( H_0 \): No effect of angle on force, and the alternative hypothesis \( H_a \): Angle affects force.

02

Determine the Degrees of Freedom

For factor A (angle), degrees of freedom \(df_A=n_A-1=3\). B (connectors), \(df_B=n_B-1=4\). Error term, \(df_E=(n_A-1)\times(n_B-1)=12\). Total degrees of freedom, \(df_T=n_A\times n_B - 1=19\).

03

Calculate Mean Squares

Mean square for factor A, \(MSA=\frac{SSA}{df_A}=\frac{58.16}{3}=19.39\). For factor B, \(MSB=\frac{SSB}{df_B}=\frac{246.97}{4}=61.7425\). For errors, \(SSE=SST-SSA-SSB=91.00\) and \(MSE=\frac{SSE}{df_E}=\frac{91.00}{12}=7.5833\).

04

Compute F-values

The F-value for factor A is \(F_A=\frac{MSA}{MSE}=\frac{19.39}{7.5833}\approx2.55\). The corresponding p-value must be evaluated using an F-distribution table (or software), with \(df_A=3\) and \(df_E=12\).

05

Compare F-value with Critical Value

The critical value at \(\alpha=0.01\) for \(df_A=3\) and \(df_E=12\) is approximately 5.95 (from F-distribution tables). Since \(F_A\approx2.55 < 5.95\), we fail to reject the null hypothesis \(H_0\).

06

Conclusion

Since the F-value for factor A (2.55) is below the critical value, there is insufficient evidence to claim the angle significantly affects the separation force at the 0.01 significance level.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a core component of inferential statistics, which allows us to make decisions or inferences about population parameters based on sample data. In the context of ANOVA (Analysis of Variance), we examine whether different groups (in this case, different angles) have different effects.

To start, we need to set up two contrasting hypotheses:

• The null hypothesis (\(H_0\)): This states that there is no effect or difference. For the exercise, it implies the angle of the pull does not affect the separation force.
• The alternative hypothesis (\(H_a\)): This suggests there is a significant difference or effect. Here, it states that the angle does indeed affect the force needed.

At a given significance level (often denoted by \(\alpha\)), usually 0.01 or 0.05, we determine if the observed data provides enough evidence to reject \(H_0\). This is done by calculating test statistics, comparing them to critical values, or analyzing p-values.

Degrees of Freedom

Degrees of freedom (df) are a crucial concept in statistics, especially in ANOVA, as they define the number of independent values or quantities which can be assigned to a statistical distribution. They are necessary to calculate mean squares and other statistics.

In the exercise, we have:

• For factor A (angles): \(df_A = n_A - 1 = 3\), where \(n_A\) is the number of levels of the angle factor.
• For factor B (connectors): \(df_B = n_B - 1 = 4\), with \(n_B\) being the number of connector samples.
• Error term: \(df_E = (n_A - 1) \times (n_B - 1) = 12\), accounting for the interaction between the levels.
• Total degrees of freedom: A combination of everything, \(df_T = n_A \times n_B - 1 = 19\).

Understanding how to allocate df in ANOVA is key because it determines how variability is partitioned among different sources.

F-distribution

The F-distribution is a probability distribution used in ANOVA to compare variances among groups, helping us detect differences in means when multiple groups are involved. It is skewed-right and defined by two types of degrees of freedom: those from the numerator and the denominator.

In an ANOVA test, as in this exercise, we compute F-values, which are ratios of variances. For example, for factor A, the F-value is calculated using: \(F_A = \frac{MSA}{MSE}\), where \(MSA\) is the mean square for factor A, and \(MSE\) is the mean square error.

By comparing this F-value to a critical value from an F-distribution table, we can decide whether to reject the null hypothesis. The critical value corresponds to our set significance level and the specific degrees of freedom for the test. If our F-value is greater than this critical value, it indicates significant variability attributable to our factor of interest.

Mean Square

Mean square is an important calculation in ANOVA, as it represents the variance estimate for a factor or an error term. It is derived by dividing the sum of squares (SS) by the corresponding degrees of freedom. There are three main mean squares in ANOVA:

• Mean square for factor A (\(MSA\)): Measures variability due to the factor, in this case, the angle of pull. In the exercise, \(MSA = \frac{SSA}{df_A} = \frac{58.16}{3} = 19.39\).
• Mean square for factor B (\(MSB\)): Reflects variance across different samples, calculated as \(MSB = \frac{SSB}{df_B}\).
• Mean square error (\(MSE\)): Accounts for variation not explained by the factors, \(MSE = \frac{SSE}{df_E} = \frac{91.00}{12} = 7.5833\).

These mean squares help calculate F-values, assessing if observed variances between group means are greater than would be expected by chance.