/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 The article "Human Lateralizatio... [FREE SOLUTION] | 91Ó°ÊÓ

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The article "Human Lateralization from Head to Foot: Sex-Related Factors" (Science, 1978: 1291-1292) reports for both a sample of righthanded men and a sample of right-handed women the number of individuals whose feet were the same size, had a bigger left than right foot (a difference of half a shoe size or more), or had a bigger right than left foot. \begin{tabular}{l|c|c|c|} & \multicolumn{1}{c}{\(\mathbf{L}>\mathbf{R}\)} & \multicolumn{1}{c}{\(\mathbf{L}=\mathbf{R}\)} & \(\mathbf{L}<\mathbf{R}\) \\ \cline { 2 - 4 } Men & 2 & 10 & 28 \\ \cline { 2 - 4 } Women & 55 & 18 & 14 \\ \cline { 2 - 4 } & & & \end{tabular} Size 40 Women 87 Does the data indicate that gender has a strong effect on the development of foot asymmetry? State the appropriate null and alternative hypotheses, compute the value of \(\chi^{2}\), and obtain information about the \(P\)-value.

Short Answer

Expert verified
Gender significantly affects foot asymmetry, as indicated by a \( P \)-value much less than 0.05.

Step by step solution

01

State the Hypotheses

To determine if gender has an effect on foot asymmetry, we set up the hypotheses as follows: - Null Hypothesis ( H_0 ): Gender is independent of foot asymmetry. - Alternative Hypothesis ( H_1 ): Gender is not independent of foot asymmetry, indicating a possible effect on development.
02

Organize the Data into a Contingency Table

Create a contingency table based on the provided data.| | \(L > R\) | \(L = R\) | \(L < R\) | Total ||--------|-----------|-----------|-----------|-------|| Men | 2 | 10 | 28 | 40 || Women | 55 | 18 | 14 | 87 || Total | 57 | 28 | 42 | 127 |This table will be used to calculate the expected frequencies and ultimately the value of \( \chi^2 \).
03

Calculate Expected Frequencies

Expected frequencies are calculated using the formula:\[E_{ij} = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}\]Calculate for each cell:- \(E_{11} = \frac{40 \times 57}{127} \approx 17.95\)- \(E_{12} = \frac{40 \times 28}{127} \approx 8.82\)- \(E_{13} = \frac{40 \times 42}{127} \approx 13.23\)- \(E_{21} = \frac{87 \times 57}{127} \approx 39.05\)- \(E_{22} = \frac{87 \times 28}{127} \approx 19.18\)- \(E_{23} = \frac{87 \times 42}{127} \approx 28.77\)
04

Compute the \( \chi^2 \) Statistic

The \( \chi^2 \) statistic is computed as:\[\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}\]Plug in the observed (O_{ij}) and expected values (E_{ij}):- \(\chi^2 = \frac{(2 - 17.95)^2}{17.95} + \frac{(10 - 8.82)^2}{8.82} + \frac{(28 - 13.23)^2}{13.23} + \)- \(\frac{(55 - 39.05)^2}{39.05} + \frac{(18 - 19.18)^2}{19.18} + \frac{(14 - 28.77)^2}{28.77}\)Calculate to get \( \chi^2 \approx 46.73 \).
05

Determine Degrees of Freedom and \( P \)-value

The degrees of freedom for a contingency table is calculated as:\[(df) = (r - 1)(c - 1)\]where \(r\) is the number of rows and \(c\) is the number of columns:\( (2-1)(3-1) = 2 \) degrees of freedom.Use a \( \chi^2 \) distribution table or calculator to find the \( P \)-value for \( \chi^2 = 46.73 \) at 2 df, which is much less than 0.05.
06

Draw a Conclusion

Since the \( P \)-value is much lower than the typical significance level of 0.05, we reject the null hypothesis. This suggests that there is a statistically significant effect of gender on foot asymmetry.

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Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there's a significant association between two categorical variables. It's particularly useful for analyzing data in the form of a contingency table, where frequencies of observed data are compared. This test calculates a statistic, known as the Chi-Square statistic (\( \chi^2 \)), which measures how much the observed frequencies deviate from expected frequencies under the null hypothesis.
  • The larger the value of \( \chi^2 \), the greater the deviation from what was expected, indicating a potential association between variables.
  • The calculation of \( \chi^2 \) involves summing up the squares of the differences between observed and expected frequencies, divided by the expected frequencies.
  • A significant Chi-Square value suggests that the association between the variables is unlikely to be due to chance.
Understanding the Chi-Square Test helps in identifying any relationships or dependencies between categorical variables.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis (\( H_0 \)) is a statement that assumes no effect or no difference in the context of the tested experiment. It's a fundamental concept that serves as a starting point for any hypothesis test.

  • The null hypothesis is typically formulated in a way to state that no association or effect exists, such as "gender is independent of foot asymmetry."
  • The goal of hypothesis testing is to determine whether there's enough evidence to reject the null hypothesis.
  • If the null hypothesis is rejected, it implies that there is a significant effect or association that warrants further investigation.
In any test like the Chi-Square test, the null hypothesis provides a baseline expectation to compare actual observed data against. Without this structured approach, it wouldn't be possible to statistically validate findings.
Gender and Foot Asymmetry
Gender and foot asymmetry investigates whether there is a correlation between a person's gender and the size differences between their left and right feet. This topic explores body asymmetry within biological contexts and the potential influence of factors like gender.

  • Foot asymmetry refers to the difference in size between a person's two feet, which can be influenced by various genetic and environmental factors.
  • When analyzing such traits, it's important to consider the distribution across differing populations, such as males and females, to understand any patterns or distinctions.
  • Statistical analysis, like the Chi-Square test, helps determine if any observed differences are statistically significant rather than random occurrences.
Understanding the potential link between gender and foot asymmetry can uncover more about human biological diversity and development.
Statistical Significance
Statistical significance is a critical concept in statistics, indicating whether the results of an analysis reflect true underlying patterns rather than random chance. It helps to decide if the evidence is strong enough to reject the null hypothesis.

  • A result is statistically significant if the p-value is below a predetermined threshold, commonly set at 0.05.
  • The p-value represents the probability that the observed data would occur under the null hypothesis's assumption.
  • In our context, if the p-value is very low, it suggests that gender likely impacts foot asymmetry beyond chance alone.
By distinguishing between random noise and true effects, statistical significance enables researchers to make informed conclusions based on their data. It serves as the benchmark for validating the findings of studies and experiments.

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Most popular questions from this chapter

Criminologists have long debated whether there is a relationship between weather conditions and the incidence of violent crime. The author of the article "Is There a Season for Homicide?" (Criminology, 1988: 287-296) classified 1361 homicides according to season, resulting in the accompanying data. Test the null hypothesis of equal proportions using \(\alpha=.01\) by using the chi-squared table to say as much as possible about the \(P\)-value. \begin{tabular}{llll} Winter & Spring & Summer & Fall \\ \hline 328 & 334 & 372 & 327 \end{tabular}

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The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002 , the top-ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3 . Let \(P_{i j}\) denote the probability that the team ranked \(i\) in its region is victorious in its game against the team ranked \(j\). Once the \(P_{i j}\) 's are available, it is possible to compute the probability that any particular seed wins its regional tournament (a complicated calculation because the number of outcomes in the sample space is quite large). The paper "Probability Models for the NCAA Regional Basketball Tournaments"(Amer. Statist., 1991: 35-38) proposed several different models for the \(P_{i j}\) 's. a. One model postulated \(P_{i j}=.5-\lambda(i-j)\) with \(\lambda=\frac{1}{32}\) (from which \(P_{16,1}=\frac{1}{32}, P_{16,2}=\frac{2}{32}\), etc.). Based on this, \(P\) (seed #1 wins) \(=.27477\), \(P(\) seed \(\\# 2\) wins \()=.20834\), and \(P\) (seed #3 wins \()=.15429\). Does this model appear to provide a good fit to the data? b. A more sophisticated model has \(P_{i j}=.5+\) \(.2813625\left(z_{i}-z_{j}\right)\), where the \(z\) 's are measures of relative strengths related to standard normal percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the range of probabilities is the same as for the model in part (a)]. The resulting probabilities of seeds 1,2 , or 3 winning their regional tournaments are \(.45883, .18813\), and \(.11032\), respectively. Assess the fit of this model.

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