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In a study of 2989 cancer deaths, the location of death (home, acute-care hospital, or chronic-care facility) and age at death were recorded, resulting in the given two-way frequency table ("Where Cancer Patients Die," Public Health Rep., 1983: 173). Using a \(.01\) significance level, test the null hypothesis that age at death and location of death are independent. $$ \begin{array}{lccc} \hline & \multicolumn{3}{c}{\text { Location }} \\ \cline { 2 - 4 } \text { Age } & \text { Home } & \text { Acute-Care } & \text { Chronic-Care } \\ \hline \mathbf{1 5 - 5 4} & 94 & 418 & 23 \\ \mathbf{5 5 - 6 4} & 116 & 524 & 34 \\ \mathbf{6 5 - 7 4} & 156 & 581 & 109 \\ \text { Over } \mathbf{7 4} & 138 & 558 & 238 \\ \hline \end{array} $$

Short Answer

Expert verified
Reject the null hypothesis; age and location of death are not independent.

Step by step solution

01

Understanding the Problem

To test the hypothesis that age and location of death are independent, we will use the Chi-Square test of independence. This involves calculating the expected frequencies assuming the null hypothesis is true, and comparing them to the observed frequencies using the Chi-Square statistic.
02

Setting Up the Hypotheses

The null hypothesis ( H_0 ) is that age at death and location of death are independent. The alternative hypothesis ( H_a ) is that age at death and location of death are not independent.
03

Calculating Row and Column Totals

Calculate the totals for each age group and each location. - Row Totals: - Age 15-54: 94 + 418 + 23 = 535 - Age 55-64: 116 + 524 + 34 = 674 - Age 65-74: 156 + 581 + 109 = 846 - Age Over 74: 138 + 558 + 238 = 934 - Column Totals: - Home: 94 + 116 + 156 + 138 = 504 - Acute-Care: 418 + 524 + 581 + 558 = 2081 - Chronic-Care: 23 + 34 + 109 + 238 = 404 - Total Deaths: 2989
04

Calculating Expected Frequencies

The expected frequency for each cell under the null hypothesis is calculated as:\[ E_{ij} = \frac{(\text{Row Total})(\text{Column Total})}{\text{Grand Total}} \]For example, for age 15-54 and Home:\[ E_{11} = \frac{(535)(504)}{2989}\approx 90.33 \]Repeat this calculation for each cell to obtain a table of expected frequencies.
05

Computing the Chi-Square Statistic

For each cell, calculate the component of the Chi-Square statistic:\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]Where \(O_{ij}\) is the observed frequency and \(E_{ij}\) is the expected frequency. Sum all the components to find the Chi-Square statistic value.
06

Determining Degrees of Freedom and Critical Value

The degrees of freedom (df) for a Chi-Square test of independence is given by:\[ df = (\text{number of rows} - 1)(\text{number of columns} - 1) \]In this case:\[ df = (4-1)(3-1) = 6 \]Using a significance level of 0.01, look up the critical value in a Chi-Square distribution table (approximately 16.812).
07

Making a Decision

Compare the calculated Chi-Square statistic to the critical value from the Chi-Square distribution table. If the statistic is greater than the critical value, reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In the Chi-Square Test of Independence, the null hypothesis ( H_0 ) is a foundational concept. It proposes that two categorical variables, such as age at death and location of death in our exercise, are independent. This means there's no association between the variables, and any observed variation is due to random chance.

When we perform the test, we want to see if indeed age at death and location are unrelated. If the calculated statistic exceeds a certain threshold (critical value), then we have enough evidence to conclude the variables are not independent, leading to the rejection of the null hypothesis. Otherwise, we "fail to reject" it, meaning we have no compelling reason to suspect a relationship between the variables.

Remember, not rejecting the null does not confirm the variables are independent; it simply indicates insufficient evidence to prove they aren't. That's a subtle, yet important distinction in hypothesis testing.
Degrees of Freedom
Degrees of freedom (df) is an essential concept when conducting a Chi-Square Test of Independence. It determines the number of values in the final calculation of a statistic that are free to vary. In a sense, it reflects the number of independent observations in your data.

For a two-way table, like in our cancer deaths study, the degrees of freedom are found by the formula: \[ df = (r - 1)(c - 1) \]where \( r \) is the number of rows and \( c \) is the number of columns. In this exercise:
  • Number of age groups (rows) = 4
  • Number of location types (columns) = 3
So, the degrees of freedom is:\[ df = (4-1)(3-1) = 6 \]

This value helps us determine which critical value to use from the Chi-Square distribution table, establishing the threshold for rejecting the null hypothesis.
Expected Frequency
Expected frequency is a key calculation in the Chi-Square Test of Independence. It represents what the frequency in each cell of a table would be if the two categories were indeed independent.

The expected frequency is calculated using this formula:\[ E_{ij} = \frac{( ext{Row Total})( ext{Column Total})}{ ext{Grand Total}} \]

For example, to find the expected number of deaths at home for the age group 15-54, you'd perform the computation:\( E_{11} = \frac{(535)(504)}{2989} \approx 90.33 \)

This value, alongside the observed frequency, is crucial for determining the chi-square statistic, since it reflects what we would theoretically see if the age of death and location are unrelated.
Observed Frequency
Observed frequency refers to the actual count of occurrences or items in each category of the table. For instance, in our example where cancer deaths are categorized, the observed frequency for the number of people aged 15-54 who died at home is 94.

These are the raw data points collected in the study, and they are compared against the expected frequencies when using the Chi-Square Test.
.br>When we compute the Chi-Square statistic, we account for the difference between observed and expected frequencies for each cell using:\[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]

This helps quantify how much the observed data deviates from what would be expected if the null hypothesis truly holds. A significant deviation indicates that the null hypothesis might not be true, suggesting a relationship between the variables.

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Most popular questions from this chapter

Say as much as you can about the \(P\)-value for an upper-tailed chi-squared test in each of the following situations: a. \(\chi^{2}=7.5, \mathrm{df}=2\) b. \(\chi^{2}=13.0, \mathrm{df}=6\) c. \(\chi^{2}=18.0, \mathrm{df}=9\) d. \(\chi^{2}=21.3, k=5\) e. \(\chi^{2}=5.0, k=4\)

Have you ever wondered whether soccer players suffer adverse effects from hitting "headers"? The authors of the article "No Evidence of Impaired Neurocognitive Performance in Collegiate Soccer Players" (Amer. J. Sports Med. 2002: 157-162) investigated this issue from several perspectives. a. The paper reported that 45 of the 91 soccer players in their sample had suffered at least one concussion, 28 of 96 nonsoccer athletes had suffered at least one concussion, and only 8 of 53 student controls had suffered at least one concussion. Analyze this data and draw appropriate conclusions. b. For the soccer players, the sample correlation coefficient calculated from the values of \(x=\) soccer exposure (total number of competitive seasons played prior to enrollment in the study) and \(y=\) score on an immediate memory recall test was \(r=-.220\). Interpret this result. c. Here is summary information on scores on a controlled oral word-association test for the soccer and nonsoccer athletes: $$ \begin{aligned} &n_{1}=26, \bar{x}_{1}=37.50, s_{1}=9.13, \\ &n_{2}=56, \bar{x}_{2}=39.63, s_{2}=10.19 \end{aligned} $$ Analyze this data and draw appropriate conclusions. d. Considering the number of prior nonsoccer concussions, the values of mean \(\pm\) SD for the three groups were soccer players, \(.30 \pm .67\); nonsoccer athletes, \(.49 \pm .87\); and student controls, .19 \(\pm .48\). Analyze this data and draw appropriate conclusions.

An information retrieval system has ten storage locations. Information has been stored with the expectation that the long-run proportion of requests for location \(i\) is given by the expression \(p_{i}=(5.5-|i-5.5|) / 30\). A sample of 200 retrieval requests gave the following frequencies for locations 1-10, respectively: \(4,15,23,25,38,31,32\), 14,10 , and 8 . Use a chi-squared test at significance level .10 to decide whether the data is consistent with the a priori proportions (use the \(P\)-value approach).

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.

A study of sterility in the fruit fly ("Hybrid Dysgenesis in Drosophila melanogaster: The Biology of Female and Male Sterility," Genetics, 1979: 161-174) reports the following data on the number of ovaries developed for each female fly in a sample of size 1,388 . One model for unilateral sterility states that each ovary develops with some probability \(p\) independently of the other ovary. Test the fit of this model using \(\chi^{2}\). \begin{tabular}{l|ccc} \(x=\) Number of Ovaries Developed & 0 & 1 & 2 \\ \hline Observed Count & 1212 & 118 & 58 \end{tabular}

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