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Chapter 11: Problem 85

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}

Short Answer

Expert verified
The missing entries are 16, 19, 28, 30, and 33. The table checks out with the initial given df=4.

Step by step solution

01

Determine Total Surplus

Given the totals in the table, we need to first calculate the only missing total, which is the fifth column (E). We know that the total for each section of columns A, B, C, D is 40. Since the overall total for each row is 100 and for columns A to D (24, 21, 12, 10), this adds up to 67. Let鈥檚 calculate the total for column E: \[ 100 - (24 + 21 + 12 + 10) = 100 - 67 = 33 \]. Therefore, column E should have a total of 33.
02

Calculate Remaining Entries

We already have a total of 40 for each column, and we need to fill the deficit. Starting with column A, which currently has 24, we find the second entry: \[ 40 - 24 = 16 \]. Similarly, for column B: \( 40 - 21 = 19 \), for column C: \( 40 - 12 = 28 \), and for column D: \( 40 - 10 = 30 \). Finally, for column E we earlier calculated the total to be 33, hence its second value is \( 33 \) (since the first entry was absent earlier).
03

Verify the Table

The table should now be completely filled with the totals for each column and row verify the consistency of our solution. - Column Inconsistencies: \( 24 + 16 = 40 \), \( 21 + 19 = 40 \), \( 12 + 28 = 40 \), \( 10 + 30 = 40 \), \( 0 + 33 = 33 \).- Row Total Consistency: The total for each cell pair matches up to 100. Therefore, our calculated table remains consistent with the total allocations of 200 and 4 degrees of freedom account correctly in balancing each column and row.
04

Calculate Degrees of Freedom

To ensure our solution is complete, confirm the degrees of freedom (df). For this contingency table with 2 rows and 5 columns, formula for degrees of freedom is: \((r-1)\times(c-1) = (2-1)\times(5-1)=1\times4=4\). This correctly corresponds to the given deduction and validation of our entries.

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Key Concepts

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

Degrees of Freedom in Contingency Tables
Understanding degrees of freedom (df) is essential when working with contingency tables. In statistics, df represents the number of values in a calculation that are free to vary. When it comes to contingency tables, this concept is crucial in determining the table's statistical validity.
  • For a contingency table with rows and columns, the degrees of freedom are calculated using the formula: diagrams.
  • In our example, the table has 2 rows and 5 columns. So, the degrees of freedom is \((2-1)\times(5-1)=4\), giving us the value of 4, which allows us to make free choices among the data points.
The degrees of freedom tell us how many pieces of independent information we can obtain from the data. Knowing that this example has a df of 4 means we need four fixed numbers to fill in the rest and complete the table. This understanding is a key concept in creating and analyzing a complete table accurately.
Cell Counts and Their Importance
Cell counts in contingency tables are the specific values within the cells that add up to the row and column totals. They serve as the foundation for the table's analysis. In order to fill in this table, you must thoroughly consider each cell count in terms of its impact on the overall totals.
  • The sum of the values in each column and row must align with their respective total counts. In our case, each column should total 40, except for column E, which totals 33 according to step-by-step calculations.
  • The accuracy of cell counts ensures the table's reliable use in any subsequent analysis, such as chi-square tests. A single incorrect cell count can disrupt the table's entirety, leading to incorrect conclusions.
In the exercise, once the totals align properly, we can confidently say that all cell counts are correctly accounted for. This ensures a logical and statistically sound foundation for more advanced analysis.
Statistical Analysis Using Contingency Tables
Contingency tables play a vital role in statistical analysis, especially in the domain of categorical data. They allow for an organized display of the frequency distribution of variables and facilitate comparative studies.
  • The purpose of the contingency table is to help you identify any potential relationships or dependencies between variables.
  • Once the table is correctly filled, it can be used to perform a chi-square test. This test helps in determining if there is a significant association between the variables summarized.
To perform a chi-square test, summative data from the cells is compared against what we would expect if no association existed. This is where the degrees of freedom directly impact the validity of the analysis, as they help determine the chi-square value's thresholds. Correct initial cell counts are critical as they form the foundation of this statistical method, ensuring the reliability of the conclusions drawn.

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Most popular questions from this chapter

Gender gap? Exercise 11.1 showed a \(2 \times 3\) table relating gender and political party identification, shown again here. The chi-squared statistic for these data equals 10.04 with a P-value of \(0.0066 .\) Conduct all five steps of the chi-squared test. \begin{tabular}{lcccc} \hline & \multicolumn{3}{c} { Political Party Identification } & \\ \cline { 2 - 4 } Gender & Democrat & Independent & Republican & Total \\ \hline Female & 421 & 398 & 244 & 1063 \\ Male & 278 & 367 & 198 & 843 \\ \hline \end{tabular}

In the GSS, subjects who were married were asked about the happiness of their marriage, the variable coded as HAPMAR. a. Go to the GSS website sda.berkeley.edu/GSS/, click GSS with no weight as the default, and construct a contingency table for 2012 relating family income (measured as in Table 11.1 ) to marital happiness: Enter FINRELA(r:4;3;2) as the row variable (where \(4=\) "above average," \(3=\) "average, \("\) and \(2=\) "below average") and HAPMAR(r:3;2;1) as the column variable \((3=\) not \(t o 0,2=\) pretty \(,\) and \(1=\) very happy \()\) As the selection filter, enter YEAR(2012). Under Output Options, put a check in the row box (instead of in the column box) for the Percentaging option and put a check in the Summary Statistics box further below. Click on Run the Table. b. Construct a table or graph that shows the conditional distributions of marital happiness, given family income. How would you describe the association? c. Compare the conditional distributions to those in Table \(11.2 .\) For a given family income, what tends to be higher, general happiness or marital happiness for those who are married? Explain.

True or false: Group 1 becomes Group 2 Interchanging two rows or interchanging two columns in a contingency table has no effect on the value of the \(X^{2}\) statistic.

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.

Down syndrome diagnostic test The table shown, from Example 8 in Chapter \(5,\) cross-tabulates whether a fetus has Down syndrome by whether the triple blood diagnostic test for Down syndrome is positive (that is, indicates that the fetus has Down syndrome). a. Tabulate the conditional distributions for the blood test result, given the true Down syndrome status. b. For the Down cases, what percentage was diagnosed as positive by the diagnostic test? For the unaffected cases, what percentage got a negative test result? Does the diagnostic test appear to be a good one? c. Construct the conditional distribution on Down syndrome status for those who have a positive test result. (Hint: You condition on the first column total and find proportions in that column.) Of those cases, what percentage truly have Down syndrome? Is the result surprising? Explain why this probability is small. \begin{tabular}{lrrr} \hline & \multicolumn{2}{c} { Blood Test Result } & \\ \cline { 2 - 3 } Down Syndrome Status & Positive & Negative & Total \\ \hline D (Down) & 48 & 6 & 54 \\ D \(^{c}\) (unaffected) & 1307 & 3921 & 5228 \\ Total & 1355 & 3927 & 5282 \\ \hline \end{tabular}
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