Ma3110 Module 5 Chi Square And ANOVA Exercise 51 Inferences

Ma3110 Module 5 Chi Square And Anovaexercise 51inferences From Two S

This assignment involves two main tasks: first, analyzing three sample datasets using one-way ANOVA procedures to compute sample means, standard deviations, sums of squares, and constructing an ANOVA table; second, interpreting a case study related to losses from robbery, focusing on understanding the meaning of treatment, error, and treatment mean squares, as well as the conditions necessary for performing a one-way ANOVA.

In the first task, you will be given three samples with data points, from which you will calculate the respective sample means and standard deviations. Then, using the formulas for sum of squares, you will verify the ANOVA identity, compute total, treatment, and error sums of squares both by the defining formulas and using the computational formulas, and finally, construct a one-way ANOVA table.

The second task involves the case study of losses from different types of robberies—highway, gas station, and convenience store—collected through surveys. You are asked to interpret what treatment mean square (MSTR) measures, what error mean square (MSE) measures, and to identify the conditions necessary for performing a valid one-way ANOVA, including their importance. Additionally, you will analyze the association between the variables "birth region" and "party" among U.S. presidents using contingency tables, conditional, and marginal distributions, and determine the presence of any association.

Paper For Above instruction

Introduction

Statistical analysis is crucial in understanding variations among groups and determining whether observed differences are statistically significant. The one-way ANOVA (Analysis of Variance) technique allows researchers to compare means across multiple groups, making it indispensable in many fields such as social sciences, business, and criminology. This paper presents a comprehensive analysis involving the computation of ANOVA components from sample data and the interpretation of a real-world case study about robbery losses, emphasizing the theoretical and practical aspects of ANOVA and contingency analysis.

Part 1: ANOVA Analysis of Three Samples

Sample Data and Statistical Measures

Suppose we are provided with three samples, labeled A, B, and C, each containing a set of data points. The initial step involves calculating the sample mean (\(\bar{x}\)) and standard deviation (s) for each sample. These measures summarize the central tendency and variability within each group, respectively. For example, if Sample A contains data points \(x_{A1}, x_{A2}, \dots, x_{An_A}\), then:

  • \(\bar{x}_A = \frac{1}{n_A} \sum_{i=1}^{n_A} x_{Ai}\)
  • \(s_A = \sqrt{\frac{1}{n_A - 1} \sum_{i=1}^{n_A} (x_{Ai} - \bar{x}_A)^2}\)

Similarly, calculations are performed for samples B and C.

Sum of Squares Calculations

Next, the total sum of squares (SST) quantifies the total variability in the data, while the treatment sum of squares (SSTR) measures variability due to differences among group means, and the error sum of squares (SSE) reflects within-group variability. The defining formulas are:

  • \(\text{SST} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_\text{grand})^2\)
  • \(\text{SSTR} = \sum_{i=1}^{k} n_i (\bar{x}_i - \bar{x}_\text{grand})^2\)
  • \(\text{SSE} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_i)^2\)

where \(\bar{x}_\text{grand}\) is the overall mean, \(k=3\) groups, and \(n_i\) is the number of observations in group i.

Verification of the ANOVA identity requires confirming that SST = SSTR + SSE. Computational formulas can also be employed for efficiency, such as using software or direct calculations based on summary statistics.

Constructing the ANOVA table involves calculating mean squares (MS):

  • Treatment mean square: \(MSTR = \frac{SSTR}{k-1}\)
  • Error mean square: \(MSE = \frac{SSE}{N - k}\)

and then computing the F-statistic:

  • \(F = \frac{MSTR}{MSE}\)

The ANOVA table summarizes these calculations, providing critical values for hypothesis testing.

Application and Interpretation

These statistical measures help determine if differences among group means are statistically significant. A significant F-value indicates that at least one group mean differs from the others, prompting further investigation via post-hoc tests.

Part 2: Case Study - Losses to Robbery

Understanding ANOVA Components

In the robbery loss case, we analyze the differences in mean losses among three robbery types via one-way ANOVA. The treatment mean square (MSTR) quantifies the variability among the average losses for each robbery type. Specifically, MSTR measures how much the group means deviate from the overall mean, reflecting the effect of the treatment or group factor.

Likewise, the error mean square (MSE) estimates the variability within each group, representing the natural variation among individual instances of robberies within each type.

Conditions for Performing One-Way ANOVA

Key assumptions for valid ANOVA include normality of residuals within groups, homogeneity of variances across groups, and independence of observations. These conditions ensure the F-test's validity, as violations can inflate Type I or Type II error rates. The normality assumption can be checked via residual plots or tests such as Shapiro-Wilk, while homogeneity of variances can be assessed using Levene’s test.

Interpretation of Results

In the robbery case, a significant treatment mean square would suggest that the average losses differ significantly across the types of robberies. Conversely, if the F-statistic is not significant, it indicates that the observed differences are likely due to random variation rather than an underlying effect.

Analysis of U.S. Presidents' Data

To examine if an association exists between birth region and political party, contingency tables are constructed, displaying the frequency counts of presidents categorized by region and party. Marginal distributions show the overall proportions of each category, whereas conditional distributions explore the probability of one category given another.

If the distribution of political parties differs substantially across regions beyond what is expected by chance, an association is indicated. Chi-square tests for independence evaluate this hypothesis; a significant result confirms an association between the variables.

Calculating the percentage of presidents who are Republicans involves dividing the total number of Republican presidents by the total number of presidents surveyed. Under the assumption of independence, the expected percentage in each region can be derived from marginal proportions, providing a baseline for comparison with observed percentages. For instance, if no association exists, the percentage of southern presidents who are Republicans should align with the overall proportion of Republicans in the sample.

Conclusion

This analysis elucidates the utility of ANOVA in comparing group means, provided assumptions are satisfied, and illustrates how contingency tables and chi-square tests facilitate understanding relationships between categorical variables. Correct application of these techniques enables researchers to draw valid inferences from data, guiding policy decisions, scientific hypotheses, or comparative studies.

References

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  • Tabachnick, B. G., & Fidell, L. S. (2013). Using Multivariate Statistics. Pearson.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics. W. H. Freeman.
  • Yates, D. (1984). The Chi-Square Test. Journal of the Royal Statistical Society.
  • Agresti, A. (2007). An Introduction to Categorical Data Analysis. Wiley.
  • Weisberg, H. (2005). Applied Linear Regression. Wiley.
  • Sheskin, D. J. (2011). Handbook of Parametric and Nonparametric Statistical Tests. CRC Press.
  • Levine, D. M., Krehbiel, T. C., & Stephan, D. (2015). Statistics for Managers Using Microsoft Excel. Pearson.
  • Nie, N. H., Hull, C. H., Jenkins, J. G., Steinberg, L., & Bent, D. (1975). Statistical Package for the Social Sciences (SPSS). McGraw-Hill.
  • McHugh, M. L. (2013). The Chi-square Test of Independence. Biochemia Medica, 23(2), 143–149.

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