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Chapter 10: Problem 24

Suppose the \(x_{i j}\) 's are "coded" by \(y_{i j}=c x_{i j}+d\). How does the value of the \(F\) statistic computed from the \(y_{i j}\) 's compare to the value computed from the \(x_{i j}\) 's? Justify your assertion.

Short Answer

Expert verified
The F statistic remains unchanged after coding.

Step by step solution

01

Understand the Coding Transformation

In this problem, the values \(x_{ij}\) are transformed to \(y_{ij}\) using the linear transformation formula \(y_{ij} = c x_{ij} + d\). This type of coding can affect calculations involving mean and variance.
02

Express the Mean Transformation

The mean of \(y_{ij}\), denoted \(\overline{y}\), is affected by the transformation as \(\overline{y} = c \overline{x} + d\), where \(\overline{x}\) is the mean of \(x_{ij}\). This indicates that the means of each set of data are linearly scaled and shifted.
03

Determine the Effect on Variance

Variance is unaffected by constant shift \(d\), but it is multiplied by \(c^2\). Therefore, the variance for \(y_{ij}\), denoted as \(\text{Var}(y)\), equals \(c^2 \text{Var}(x)\).
04

Analyze the Effect on Within-Group and Between-Group Variance

Both the within-group variance \(S_W^2\) and the between-group variance \(S_B^2\) are scaled by \(c^2\) due to the linear transformation. \(S_W^2(y) = c^2 S_W^2(x)\) and \(S_B^2(y) = c^2 S_B^2(x)\).
05

Calculate the F-Statistic for y_{ij}

The \(F\) statistic ratio can be formulated as \(F = \frac{S_B^2}{S_W^2}\). In this case, \(F_y = \frac{c^2 S_B^2(x)}{c^2 S_W^2(x)} = \frac{S_B^2(x)}{S_W^2(x)} = F_x\).
06

Conclude F-Statistic Comparison

Since \(c^2\) cancels out in both the numerator and the denominator of the \(F\) statistic calculation for \(y_{ij}\), the \(F\) statistic remains unchanged from \(x_{ij}\) to \(y_{ij}\).

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Key Concepts

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

Understanding the F-statistic
The F-statistic is a crucial component in ANOVA (Analysis of Variance) that helps us test if there are significant differences between group means. It compares the between-group variance and the within-group variance.
This ratio, expressed as \( F = \frac{S_B^2}{S_W^2} \), helps us decide if the variation among the sample means is more than would be expected by random chance alone.
A higher F-statistic often indicates a more substantial difference between group means.
Linear Transformation and Its Impacts
A linear transformation involves scaling and shifting data in a systematic way. The formula \( y_{ij} = c x_{ij} + d \) is used for linear transformation:
  • \( c \) is the scaling factor
  • \( d \) is the shifting term

The transformation affects mean and variance differently:
  • The mean gets scaled by \( c \) and shifted by \( d \).
  • The variance is scaled by \( c^2 \), but remains unaffected by \( d \).
Mean Effect in Linear Transformation
The mean of a transformed dataset is essentially a linear function of the original mean.
Using the equation \( \overline{y} = c \overline{x} + d \), we can see how the transformation changes the mean:
  • \( c \overline{x} \) scales the mean of the original dataset, \( \overline{x} \).
  • \( d \) adds a shift, moving the mean up or down.

Thus, while transforming affects the mean, it is essential to note that these changes do not impact the F-statistic itself.
Variance Effect in Linear Transformation
Variance, a measure of how much values in a dataset differ from the mean, is altered by the scaling factor \( c \) in the transformation \( y_{ij} = c x_{ij} + d \).
  • The constant \( d \) does not affect variance because variance measures spread, not location.
  • The scaling factor \( c \) is crucial because it affects variance by a factor of \( c^2 \).

For transformed data, variance becomes \( \text{Var}(y) = c^2 \text{Var}(x) \).
Interestingly, since both within-group and between-group variances get multiplied by \( c^2 \), the F-statistic ratio remains unchanged.

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Most popular questions from this chapter

An experiment was carried out to compare flow rates for four different types of nozzle. a. Sample sizes were \(5,6,7\), and 6, respectively, and calculations gave \(f=3.68\). State and test the relevant hypotheses using \(\alpha=.01\) b. Analysis of the data using a statistical computer package yielded \(P\)-value \(=.029\). At level \(.01\), what would you conclude, and why?

The article "Computer-Assisted Instruction Augmented with Planned Teacher/Student Contacts" (J. of Exp. Educ., 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-hour exam. In addition, a 10-minute retention test was administered 6 weeks later. Summary quantities are given. \begin{tabular}{lcccc} & \multicolumn{2}{c}{ Exam } & \multicolumn{2}{c}{ Retention Test } \\ \hline Method & \(\bar{x}_{\digamma}\) & \(s_{i}\) & \(\bar{x}_{k}\) & \(s_{i}\) \\ \hline L/D & \(29.3\) & \(4.99\) & \(30.20\) & \(3.82\) \\ \(\mathrm{R}\) & \(28.0\) & \(5.33\) & \(28.80\) & \(5.26\) \\ \(\mathrm{R} / \mathrm{L}\) & \(30.2\) & \(3.33\) & \(26.20\) & \(4.66\) \\ \(\mathrm{C}\) & \(32.4\) & \(2.94\) & \(31.10\) & \(4.91\) \\ \(\mathrm{C} / \mathrm{L}\) & \(34.2\) & \(2.74\) & \(30.20\) & \(3.53\) \\ \hline \end{tabular} 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.

In an experiment to compare the quality of four different brands of magnetic recording tape, five 2400 -ft reels of each brand \((A-D)\) were selected and the number of flaws in each reel was determined. A: \(10 \quad 5 \quad 12 \quad 14 \quad 8\) B: \(\begin{array}{llllll}14 & 12 & 17 & 9 & 8\end{array}\) C: \(13 \quad 18 \quad 10 \quad 15 \quad 18\) D: \(\begin{array}{lllll}17 & 16 & 12 & 22 & 14\end{array}\) It is believed that the number of flaws has approximately a Poisson distribution for each brand. Analyze the data at level .01 to see whether the expected number of flaws per reel is the same for each brand.

The article "Origin of Precambrian Iron Formations" (Econ. Geology, 1964: 1025-1057) reports the following data on total Fe for four types of iron formation ( \(1=\) carbonate, \(2=\) silicate, \(3=\) magnetite, \(4=\) hematite). \(\begin{array}{llllll}\text { 1: } & 20.5 & 28.1 & 27.8 & 27.0 & 28.0 \\ & 25.2 & 25.3 & 27.1 & 20.5 & 31.3 \\ \text { 2: } & 26.3 & 24.0 & 26.2 & 20.2 & 23.7 \\ & 34.0 & 17.1 & 26.8 & 23.7 & 24.9 \\ 3: & 29.5 & 34.0 & 27.5 & 29.4 & 27.9 \\ & 26.2 & 29.9 & 29.5 & 30.0 & 35.6 \\ 4: & 36.5 & 44.2 & 34.1 & 30.3 & 31.4 \\ & 33.1 & 34.1 & 32.9 & 36.3 & 25.5\end{array}\) Carry out an analysis of variance \(F\) test at significance level \(.01\), and summarize the results in an ANOVA table.

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. \(\begin{array}{llll}\text { 1: } & 85.06 & 85.25 & 84.87 \\ \text { 2: } & 84.99 & 84.28 & 84.88 \\ \text { 3: } & 84.48 & 84.72 & 85.10 \\ \text { 4: } & 84.10 & 84.55 & 84.05\end{array}\)
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