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Chapter 11: Problem 35

The number of miles of useful tread wear (in 1000's) was determined for tires of each of five different makes of subcompact car (factor \(A\), with \(I=5\) ) in combination with each of four different brands of radial tires (factor \(B\), with \(J=4\) ), resulting in \(I J=20\) observations. The values \(\mathrm{SSA}=30.6, \mathrm{SSB}=44.1\), and \(\mathrm{SSE}=\) \(59.2\) were then computed. Assume that an additive model is appropriate. a. Test \(H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha_{4}=\alpha_{5}=0\) (no differences in true average tire lifetime due to makes of cars) versus \(H_{\mathrm{a}}\) : at least one \(\alpha_{i} \neq 0\) using a level \(.05\) test. b. \(H_{0}: \beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\) (no differences in true average tire lifetime due to brands of tires) versus \(H_{\mathrm{a}}\) : at least one \(\beta_{j} \neq 0\) using a level \(.05\) test.

Short Answer

Expert verified
a. Fail to reject \(H_0\). b. Reject \(H_0\).

Step by step solution

01

Determine Total Sum of Squares (SST)

First, we compute the total sum of squares, \[ \text{SST} = \text{SSA} + \text{SSB} + \text{SSE}. \]Substitute the given values: \[\text{SST} = 30.6 + 44.1 + 59.2 = 133.9. \]
02

Calculate Degrees of Freedom

Calculate the degrees of freedom for each source of variation.- For factor A (car makes): \(df_A = I - 1 = 5 - 1 = 4\).- For factor B (tire brands): \(df_B = J - 1 = 4 - 1 = 3\).- Error: \(df_E = IJ(n-1)\), assuming \(n = 1\) since specific number of repetitions is not given, \(df_E = 20 - 5 - 4 = 15\).- Total: \(df_T = IJ - 1 = 20 - 1 = 19\).
03

Calculate Mean Squares

Mean Square for factor A (MSA): \[ \text{MSA} = \frac{\text{SSA}}{df_A} = \frac{30.6}{4} = 7.65.\]Mean Square for factor B (MSB):\[ \text{MSB} = \frac{\text{SSB}}{df_B} = \frac{44.1}{3} = 14.7.\]Mean Square for Error (MSE):\[ \text{MSE} = \frac{\text{SSE}}{df_E} = \frac{59.2}{15} = 3.9467.\]
04

Conduct F-test for Factor A

Compute the F-statistic for factor A:\[ F_A = \frac{\text{MSA}}{\text{MSE}} = \frac{7.65}{3.9467} \approx 1.94.\]Compare this value to the \( F \)-critical value for \( df_{1}=4 \) and \( df_{2}=15 \) at \( \alpha = 0.05 \). The critical value from the F-distribution table is approximately 3.06.Since \( F_A =1.94 < 3.06 \), we fail to reject the null hypothesis \( H_0: \alpha_1 = \alpha_2 = \ldots = \alpha_5 = 0 \).
05

Conduct F-test for Factor B

Compute the F-statistic for factor B:\[ F_B = \frac{\text{MSB}}{\text{MSE}} = \frac{14.7}{3.9467} \approx 3.726.\]Compare this value to the \( F \)-critical value for \( df_{1}=3 \) and \( df_{2}=15 \) at \( \alpha = 0.05 \). The critical value from the F-distribution table is around 3.29.Since \( F_B =3.726 > 3.29 \), we reject the null hypothesis \( H_0: \beta_1 = \beta_2 = \beta_3 = \beta_4 = 0 \).

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Key Concepts

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

Factorial Experiment
A factorial experiment involves studying the effects of two or more factors simultaneously. The aim is to understand how these independent variables interact with each other and influence the dependent variable. In our tire wear example, we have factors such as car make and tire brand. Both are being examined for their impact on the lifespan of a tire.

A factorial experiment setup is different from a simple one-variable design because it allows us to observe not only the main effects of each factor but also their interactions.
  • **Main effects:** These are the individual effects of each factor. For instance, how does the car make alone impact tire wear?
  • **Interaction effects:** This checks whether there's a combined effect when factors like car make and tire brand take action together. This might mean that one specific tire brand performs better or worse on certain car makes.
Factorial experiments save time and resources by allowing for more comprehensive data collection in one experiment rather than multiple separate ones. They provide a full picture of how all factors work separately and in conjunction.
Hypothesis Testing
Hypothesis testing in statistics is a way to decide if there's enough evidence to reject a statement about a population parameter. It's like a courtroom trial, where we assess evidence before reaching a verdict.
In our problem, we're testing two hypotheses for factor A (car makes) and factor B (tire brands).

**For Factor A:**
  • **Null Hypothesis (\( H_0 \)):** Assumes there is no difference in tire longevity across car makes.
  • **Alternative Hypothesis (\( H_a \)):** At least one car make has a different tire longevity.
**For Factor B:**
  • **Null Hypothesis (\( H_0 \)):** Assumes there is no difference in tire longevity across tire brands.
  • **Alternative Hypothesis (\( H_a \)):** At least one tire brand affects tire longevity differently.
In hypothesis testing, we use statistical calculations to determine whether to reject the null hypothesis or not. We base the decision on observed data and predefined significance levels such as 0.05, which gives us a structured way to make conclusions based on probability.
F-distribution
The F-distribution is an essential concept in performing ANOVA (Analysis of Variance) tests. It is a probability distribution that is used to compare two variances and gauge the statistical significance of the observed differences among group means.

In the tire wear study, we used the F-distribution to determine whether the variances associated with the car makes and tire brands were statistically different from one another.

**Key Features:**
  • **Shape:** The shape of the F-distribution varies with degrees of freedom. It is positively skewed, meaning it extends more broadly on the right.
  • **Degrees of Freedom:** These are based on the number of groups and sample sizes. For our experiment, we calculated degrees for both factors and total errors, crucial for identifying the critical value in the F-table.
  • **Critical Values:** To make decisions in hypothesis testing, we compare calculated F-values to critical ones. Determining these values involves using an F-table or statistical software, considering our significance level and degrees of freedom.
This distribution is instrumental for checking if observed differences in the data are meaningful or could just be due to random chance. Therefore, grasping the F-distribution helps in effective decision-making through ANOVA.

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Most popular questions from this chapter

The article "Computer-Assisted Instruction Augmented with Planned Teacher/Student Contacts" (J. Exper. Ed., Winter 1980-1981: 120-126) compared five different methods for teaching descriptive statistics. The five methods were traditional lecture and discussion (L/D), programmed textbook instruction (R), programmed text with lectures (R/L), computer instruction (C), and computer instruction with lectures (C/L). Forty-five students were randomly assigned, 9 to each method. After completing the course, the students took a 1-h exam. In addition, a 10-minute retention test was administered 6 weeks later. Summary quantities are given. $$ \begin{array}{lcccc} & {\text { Exam }} & {\text { Retention Test }} \\ \text { Method } & \overline{\boldsymbol{x}}_{\boldsymbol{i}} \cdot & \boldsymbol{s}_{\boldsymbol{i}} & \overline{\boldsymbol{x}}_{\boldsymbol{i}} \cdot & \boldsymbol{s}_{\boldsymbol{i}} \\ \hline \text { L/D } & 29.3 & 4.99 & 30.20 & 3.82 \\ \text { R } & 28.0 & 5.33 & 28.80 & 5.26 \\ \text { R/L } & 30.2 & 3.33 & 26.20 & 4.66 \\ \text { C } & 32.4 & 2.94 & 31.10 & 4.91 \\ \text { C/L } & 34.2 & 2.74 & 30.20 & 3.53 \end{array} $$ The grand mean for the exam was \(30.82\), and the grand mean for the retention test was \(29.30\). a. Does the data suggest that there is a difference among the five teaching methods with respect to true mean exam score? Use \(\alpha=.05\). b. Using a \(.05\) significance level, test the null hypothesis of no difference among the true mean retention test scores for the five different teaching methods.

When both factors are random in a two-way ANOVA experiment with \(K\) replications per combination of factor levels, the expected mean squares are \(E(\mathrm{MSE})=\sigma^{2}, E(\mathrm{MSA})=\sigma^{2}+\) \(K \sigma_{G}^{2}+J K \sigma_{A}^{2}, E(\mathrm{MSB})=\sigma^{2}+K \sigma_{G}^{2}+I K \sigma_{B}^{2}\), and \(E(\mathrm{MSAB})=\sigma^{2}+K \sigma_{G}^{2}\) a. What \(F\) ratio is appropriate for testing \(H_{0 G}: \sigma_{G}^{2}=0\) versus \(H_{\mathrm{a} G}: \sigma_{G}^{2}>0\) ? b. Answer part (a) for testing \(H_{0 A}: \sigma_{A}^{2}=0\) versus \(H_{\mathrm{aA}}: \sigma_{A}^{2}>0\) and \(H_{0 B}: \sigma_{B}^{2}=0\) versus \(H_{\mathrm{a} B}: \sigma_{B}^{2}>0\)

Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in \(.0001\) in.) is determined. The depths are shown in this table: $$ \begin{aligned} &\text { Soil Type (B) }\\\ &\begin{array}{l|lll} & \mathbf{1} & \mathbf{2} & \mathbf{3} \\ \hline \mathbf{1} & 64 & 49 & 50 \\ \mathbf{2} & 53 & 51 & 48 \\ \mathbf{3} & 47 & 45 & 50 \\ \mathbf{4} & 51 & 43 & 52 \\ \hline \end{array} \end{aligned} $$ a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use \(\alpha=.05\). b. Compute \(\hat{\mu}, \hat{\alpha}_{1}, \hat{\alpha}_{2}, \hat{\alpha}_{3}, \hat{\alpha}_{4}, \hat{\beta}_{1}, \hat{\beta}_{2}\), and \(\hat{\beta}_{3}\)

Four laboratories \((1-4)\) are randomly selected from a large population, and each is asked to make three determinations of the percentage of methyl alcohol in specimens of a compound taken from a single batch. Based on the accompanying data, are differences among laboratories a source of variation in the percentage of methyl alcohol? State and test the relevant hypotheses using significance level .05. \(\begin{array}{llll}1: & 85.06 & 85.25 & 84.87 \\ 2: & 84.99 & 84.28 & 84.88 \\\ 3: & 84.48 & 84.72 & 85.10 \\ 4: & 84.10 & 84.55 & 84.05\end{array}\)

a. Show that a constant \(d\) can be added to (or subtracted from) each \(x_{i j}\) without affecting any of the ANOVA sums of squares. b. Suppose that each \(x_{i j}\) is multiplied by a nonzero constant \(c\). How does this affect the ANOVA sums of squares? How does this affect the values of the \(F\) statistics \(F_{A}\) and \(F_{B}\) ? What effect does "coding" the data by \(y_{i j}=c x_{i j}+d\) have on the conclusions resulting from the ANOVA procedures?
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