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Babies and gray hair A young child wonders what causes women to have babies. For each woman who lives on her block, she observes whether her hair is gray and whether she has young children, with the results shown in the table that follows. a. Construct the \(2 \times 2\) contingency table that cross-tabulates gray hair (yes, no) with has young children (yes, no) for these nine women. b. Treating has young children as the response variable, obtain the conditional distributions for those women who have gray hair and for those who do not. Does there seem to be an association? c. Noticing this association, the child concludes that not having gray hair is what causes women to have children. Use this example to explain why association does not necessarily imply causation. \begin{tabular}{lcc} \hline Woman & Gray Hair & Young Children \\ \hline Andrea & No & Yes \\ Mary & Yes & No \\ Linda & No & Yes \\ Jane & No & Yes \\ Maureen & Yes & No \\ Judy & Yes & No \\ Margo & No & Yes \\ Carol & Yes & No \\ Donna & No & Yes \\ \hline \end{tabular}

Short Answer

Expert verified
Constructed a table and calculated probabilities revealed an association, but correlation does not equal causation; age or other factors could influence both variables.

Step by step solution

01

Understand Data Types

The table shows the attributes of nine women, identifying whether they have gray hair ('Yes' or 'No') and if they have young children ('Yes' or 'No'). We will use these attributes to construct a contingency table.
02

Construct Contingency Table

Build a 2x2 contingency table utilizing the data provided. \[\begin{array}{cc|c}\text{Gray Hair} & \text{Young Children} & \text{Total} \\hline\text{Yes} & \text{Yes} & 0 \\text{Yes} & \text{No} & 4 \\hline\text{No} & \text{Yes} & 5 \\text{No} & \text{No} & 0 \\end{array}\]This table organizes the observations regarding hair color and the presence of young children.
03

Calculate Conditional Distributions

For women with gray hair: - P(Young Children | Gray Hair) = \( \frac{0}{4} = 0 \) - P(No Young Children | Gray Hair) = \( \frac{4}{4} = 1 \) For women without gray hair: - P(Young Children | No Gray Hair) = \( \frac{5}{5} = 1 \) - P(No Young Children | No Gray Hair) = \( \frac{0}{5} = 0 \) These probabilities illustrate the relationships between hair color and having young children.
04

Analyze Association

Compare the conditional probabilities from Step 3: Among women with gray hair, none have young children. Among women without gray hair, all have young children. This suggests an association between not having gray hair and having young children.
05

Discuss Association vs. Causation

The association found between having no gray hair and having children does not imply causation. There could be underlying variables such as age affecting both hair color and likelihood of having children. Correlation does not inherently imply that one factor causes the other.

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Key Concepts

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

Conditional Distribution
A conditional distribution helps us understand probabilities within a sub-group of a population. Here, we focus on women with and without gray hair. A conditional distribution looks at the likelihood of having young children within these sub-groups.
In our exercise, we calculated these probabilities:
  • For women with gray hair, the probability of having young children is 0 (since no woman in this group has young children).
  • For women without gray hair, this probability is 1 (since all women in this group have young children).
Exploring Conditional Distributions: - The conditional probability for a sub-group is calculated by dividing the number of favorable outcomes by the total number of cases in that sub-group.
Knowing these distributions helps identify patterns, like whether one group is more likely to have a certain characteristic compared to another.
Association vs. Causation
It is essential to understand that just because two variables are associated does not mean one causes the other. In our case, there is an observed association between not having gray hair and having young children. This association might lead one to believe that hair color affects childbearing.
However, this is where the term 'causation' comes in. Causation means one event is the result of the occurrence of the other event. In scientific terms, to prove causation, we often need more complex analyses and controlled experiments. Here are some reasons why association does not imply causation:
  • There could be a third variable, such as age, influencing both factors.
  • Sometimes, associations are merely coincidences.
  • The time sequence between events hasn't been established to infer causality.
To determine causation, it would require longitudinal studies or controlled experiments to definitively show that changing one variable directly alters another.
Data Analysis
Analyzing data involves processing and interpreting information to uncover patterns or truths about a problem. In the given problem, we constructed a contingency table. This table is crucial for visualizing the data, as it summarizes the relationship between two categorical variables.
Data analysis methods include:
  • Constructing tables like these to clarify relationships between variables.
  • Calculating probabilities, such as conditional probabilities, to understand the likelihood of various outcomes within groups.
  • Interpreting these probabilities to determine whether there's potential association between the variables.
Why Data Analysis Matters:
- It helps inform decisions based on statistical evidence. - It enables us to make sense of raw data, moving from simple observation to actionable insight. - Analyzing data correctly allows for well-founded conclusions, aiding knowledge-building and problem-solving.

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Most popular questions from this chapter

In a study conducted by a pharmaceutical company, 605 out of 790 smokers and 122 out of 434 nonsmokers were diagnosed with lung cancer. a. Construct a \(2 \times 2\) contingency table relating smoking (SMOKING, categories smoker and nonsmoker) as the rows to lung cancer (LUNGCANCER, categories present and absent) as the columns. b. Find the four expected cell counts when assuming independence. Compare them to the observed cell counts, identifying cells having more observations than expected. c. For this data, \(X^{2}=272.89 .\) Verify this value by plugging into the formula for \(X^{2}\) and computing the sum.

Standardized residuals for \(2 \times 2\) tables The table that follows shows the standardized residuals in parentheses for GSS data about the statement, "Women should take care of running their homes and leave running the country up to men." The absolute value of the standardized residual is 13.2 in every cell. For chi-squared tests with \(2 \times 2\) tables, since \(d f=1,\) only one nonredundant piece of information exists about whether an association exists. If observed count \(\geq\) expected count in one cell, observed count \(<\) expected count in the other cell in that row or column. Explain why this is true, using the fact that observed and expected counts have the same row and column totals. (In fact, in \(2 \times 2\) tables, all four standardized residuals have absolute value equal to the square root of the \(X^{2}\) test statistic.) \begin{tabular}{lcc} \hline Year & Agree & Disagree \\ \hline 1974 & 509(13.2) & 924(-13.2) \\ 1998 & 280(-13.2) & 1534(13.2) \\ \hline \end{tabular}

Statistical versus practical significance In any significance test, when the sample size is very large, we have not neces sarily established an important result when we obtain statistical significance. Explain what this means in the context of analyzing contingency tables with a chi-squared test.

Degrees of freedom explained For testing independence in a contingency table of size \(r \times c,\) the degrees of freedom (df) for the chi-squared distribution equal \(d f=(r-1) \times(c-1) .\) They have the following interpretation: Given the row and column marginal totals in an \(r \times\) contingency table, the cell counts in a rectangular block of size \((r-1) \times(c-1)\) determine all the other cell counts. Consider the following table, which cross-classifies political views by whether the subject would ever vote for a female president, based on the 2010 GSS. For this \(3 \times 2\) table, suppose we know the counts in the upper left-hand \((3-1) \times(2-1)=2 \times 1\) block of the table, as shown. \begin{tabular}{lccc} \hline & \multicolumn{2}{c} { Vote for Female } & \\ & \multicolumn{2}{c} { President } & \\ \cline { 2 - 3 } Political Views & Yes & No & Total \\ \hline Extremely Liberal & 56 & & 58 \\ Moderate & 490 & & 509 \\ Extremely Conservative & & & 61 \\ \hline Total & 604 & 24 & 628 \\ \hline \end{tabular} a. Given the cell counts and the row and column totak, fill in the counts that must appear in the blank cells. b. Now, suppose instead of the preceding table, you are shown the following table, this time only revealing a \(2 \times 1\) block in the lower-right part. Find the counts in the remaining cells. \begin{tabular}{lccc} \hline & \multicolumn{2}{c} { Vote for Female } & \\ & \multicolumn{2}{c} { President } & \\ \cline { 2 - 3 } Political Views & Yes & No & Total \\ \hline Extremely Liberal & & & \\ Moderate & & 58 \\ Extremely Conservative & & 3 & 61 \\ \hline Total & & 19 & 509 \\ \hline \end{tabular} This example serves to show that once the marginal totals are fixed in a contingency table, a block of only cell counts is free to vary. Once \((r-1) \times(c-1)\) these are given (as in part a or \(\mathrm{b}\) ), the remaining cell counts follow automatically. The value for the degrees of freedom is exactly the number of cells in this block, \(d f=(r-1) \times(c-1)\) or

What is \(d f\) ? The contingency table that follows has \(d f=4\). Show that you can reconstruct the entire table by filling in the remaining cell counts based on the 4 cell counts shown. \begin{tabular}{lllllll} \hline & \(\mathrm{A}\) & \(\mathrm{B}\) & \(\mathrm{C}\) & \(\mathrm{D}\) & \(\mathrm{E}\) & Total \\ & 24 & 21 & 12 & 10 & \(-\) & 100 \\ & \(-\) & \(-\) & \(-\) & \(-\) & \(-\) & 100 \\ Total & 40 & 40 & 40 & 40 & 40 & 200 \\ \hline \end{tabular}

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