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We give sample sizes for the groups in a dataset and an outline of an analysis of variance table with some information on the sums of squares. Fill in the missing parts of the table. What is the value of the F-test statistic? Four groups with \(n_{1}=10, n_{2}=10, n_{3}=10\), and \(n_{4}=10 .\) ANOVA table includes: $$ \begin{array}{|l|l|l|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } \\ \hline \text { Groups } & & 960 & & \\ \hline \text { Error } & & 5760 & & \\ \hline \text { Total } & & 6720 & & \\ \hline \end{array} $$

Short Answer

Expert verified
So, the missing parts are \(df_{groups} = 3\), \(df_{errors} = 36\), \(MS_{groups} = 320\), \(MS_{errors} = 160\), and the F-statistic is 2.

Step by step solution

01

Calculate Degree of Freedom (df)

Calculate the degrees of freedom for the groups and for the errors. Degree of freedom for groups is given by \(df_{groups} = k - 1\) where \(k\) is the number of groups (in this case, 4). Hence \(df_{groups} = 4 - 1 = 3\). On the other hand, the degree of freedom for errors is given by \(df_{errors} = N - k\), where \(N\) is the total sample size and \(k\) is the number of groups. Hence \(df_{errors} = 40 - 4 = 36\).
02

Calculate Mean Square (MS)

We derive Mean Square (MS) for groups and errors by dividing the Sum of Squares (SS) by the respective degrees of freedom. Therefore, \(MS_{groups} = SS_{groups} / df_{groups} = 960 / 3 = 320\) and \(MS_{errors} = SS_{errors} / df_{errors} = 5760 / 36 = 160\).
03

Compute the F-statistic

The F-statistic is given by the ratio of the Mean Square of the groups to the Mean Square of the errors. Applying the obtained values, we have \(F-statistic = MS_{groups} / MS_{errors} = 320 / 160 = 2\).

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Key Concepts

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

Degrees of Freedom
When working with statistical analyses, such as the Analysis of Variance (ANOVA), the term 'degrees of freedom' (df) is an essential concept. It refers to the number of independent pieces of information available to estimate another statistic.

In the context of ANOVA, the degrees of freedom for a group, often notated as \( df_{groups} \), is calculated by taking the number of groups and subtracting one, \( df_{groups} = k - 1 \), where \( k \) represents the number of groups. With four groups in our exercise, we have \( df_{groups} = 4 - 1 = 3 \). This represents the number of independent comparisons that can be made between group means.

Similarly, the degrees of freedom for errors, \( df_{errors} \), is found by subtracting the number of groups from the total sample size, \( df_{errors} = N - k \), where \( N \) is the total number of observations. In our case, with 40 observations across all groups, \( df_{errors} = 40 - 4 = 36 \). These degrees of freedom reflect the variability within the groups.

Understanding degrees of freedom is crucial as it affects the calculation of statistical tests and the interpretation of results.
Mean Square (MS)
Mean Square (MS) is another key term used in ANOVA to measure the variance within the groups (\( MS_{groups} \)) and the variance within the errors (\( MS_{errors} \)).

To calculate MS, you divide the sum of squares (SS) for each source by its corresponding degrees of freedom. For instance, \( MS_{groups} = SS_{groups} / df_{groups} \). In the exercise, the SS for the groups is 960 and, as computed previously, \( df_{groups} \) is 3, hence \( MS_{groups} = 960 / 3 = 320 \). This value represents the average variation between the means of the groups.

Similarly, \( MS_{errors} \) is calculated by dividing the SS for errors by \( df_{errors} \). With the SS for errors at 5760 and 36 degrees of freedom, \( MS_{errors} = 5760 / 36 = 160 \), indicating the average variation within the groups. The MS measures are critical for comparison in the ANOVA to understand whether the differences observed between group means are significant or not.
F-test Statistic
The F-test statistic is a crucial output of ANOVA and is utilized to determine whether there are any statistically significant differences between the means of different groups. It is calculated as the ratio of the Mean Square for groups (\( MS_{groups} \)) to the Mean Square for errors (\( MS_{errors} \)).

By dividing the variance between the groups by the variance within the groups, the F-test statistic, \( F = MS_{groups} / MS_{errors} \), gives us a measure of between-group variability relative to the within-group variability. For our exercise, with the calculated \( MS_{groups} = 320 \) and \( MS_{errors} = 160 \), the F-test statistic would be \( F = 320 / 160 = 2 \).

An F-statistic value greater than 1 suggests a larger between-group variance compared to within-group variance, which is a signal of potential differences between group means. Ultimately, this statistic will be compared against a critical value from an F-distribution, considering the degrees of freedom, to decide if the observed differences are significant.

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Most popular questions from this chapter

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) What is the pooled standard deviation? What degrees of freedom are used in doing inferences for these means and differences in means?

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(90 \%\) confidence interval for the difference in the means of populations \(\mathrm{B}\) and \(\mathrm{C}\).

Exercises 8.41 to 8.45 refer to the data with analysis shown in the following computer output: \(\begin{array}{lrrr}\text { Level } & \text { N } & \text { Mean } & \text { StDev } \\ \text { A } & 5 & 10.200 & 2.864 \\ \text { B } & 5 & 16.800 & 2.168 \\ \text { C } & 5 & 10.800 & 2.387\end{array}\) \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Groups } & 2 & 133.20 & 66.60 & 10.74 & 0.002 \\ \text { Error } & 12 & 74.40 & 6.20 & & \\ \text { Total } & 14 & 207.60 & & & \end{array}\) Find a \(95 \%\) confidence interval for the mean of population \(\mathrm{A}\).

Data 4.1 introduces a study in mice showing that even low-level light at night can interfere with normal eating and sleeping cycles. In the full study, mice were randomly assigned to live in one of three light conditions: LD had a standard light/dark cycle, LL had bright light all the time, and DM had dim light when there normally would have been darkness. Exercises 8.28 to 8.34 in Section 8.1 show that the groups had significantly different weight gain and time of calorie consumption. In Exercises 8.55 and \(8.56,\) we revisit these significant differences. Body Mass Gain Computer output showing body mass gain (in grams) for the mice after four weeks in each of the three light conditions is shown, along with the relevant ANOVA output. Which light conditions give significantly different mean body mass gain? \(\begin{array}{lrrr}\text { Level } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { DM } & 10 & 7.859 & 3.009 \\ \text { LD } & 9 & 5.987 & 1.786 \\ \text { LL } & 9 & 11.010 & 2.624\end{array}\) One-way ANOVA: BM4Gain versus Light \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\ \text { Light } & 2 & 116.18 & 58.09 & 8.96 & 0.001 \\ \text { Error } & 25 & 162.10 & 6.48 & & \\ \text { Total } & 27 & 278.28 & & & \end{array}\)

Two sets of sample data, \(\mathrm{A}\) and \(\mathrm{B}\), are given. Without doing any calculations, indicate in which set of sample data, \(\mathrm{A}\) or \(\mathrm{B}\), there is likely to be stronger evidence of a difference in the two population means. Give a brief reason, comparing means and variability, for your answer. $$ \begin{array}{cc|cc} \hline {\text { Dataset A }} & {\text { Dataset B }} \\ \hline \text { Group 1 } & \text { Group 2 } & \text { Group 1 } & \text { Group 2 } \\ \hline 13 & 18 & 13 & 48 \\ 14 & 19 & 14 & 49 \\ 15 & 20 & 15 & 50 \\ 16 & 21 & 16 & 51 \\ 17 & 22 & 17 & 52 \\ \bar{x}_{1}=15 & \bar{x}_{2}=20 & \bar{x}_{1}=15 & \bar{x}_{2}=50 \end{array} $$

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