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Chapter 10: Problem 11

According to the website MedicalNewsToday.com, coronary artery disease accounts for about \(40 \%\) of deaths in the United States. Many people believe this is due to modern-day factors such as high-calorie fast food and lack of exercise. However, a study published in the Journal of the American Medical Association in November 2009 (www.medicalnewstoday .com) reported on 16 mummies from the Egyptian National Museum of Antiquities in Cairo. The mummies were examined, and 9 of them had hardening of the arteries, which seems to suggest that hardening of the arteries is not a new problem. a. Calculate the expected number of mummies with artery disease (assuming the rate is the same as in the modern day). Then calculate the expected number of mummies without artery disease (the rest). b. Calculate the observed value of the chi-square statistic for these mummies.

Short Answer

Expert verified
The expected number of mummies with artery disease is 6.4 and without artery disease is 9.6. The observed value of the chi-square statistic for these mummies is 1.72

Step by step solution

01

Calculate Expected Values

First, calculate the expected number of mummies with artery disease. This is found by multiplying the total number of mummies by the proportion of deaths due to coronary artery disease in modern times. Let's denote number of mummies as \( N \) (N=16), the proportion of modern population with artery disease as \( P \) (P=0.40)The expected number of mummies with artery disease is \( N \times P = 16 \times 0.40 = 6.4 \)Next, calculate the expected number of mummies without artery disease. This is found by subtracting the expected number with artery disease from the total number of mummies. Expected number of mummies without artery disease = \( N - N \times P = 16 - 6.4 = 9.6 \)
02

Calculate the Chi-Square Statistic

The chi-square statistic is calculated using the formula \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \( O_i \) represents the observed frequency and \( E_i \) is the expected frequencyFor mummies with artery disease: \((O_1 - E_1)^2 / E_1 = (9 - 6.4)^2 / 6.4 = 1.04\)For mummies without artery disease: \((O_2 - E_2)^2 / E_2 = (7 - 9.6)^2 / 9.6 = 0.68\)So, the value of the chi-square statistic would be the sum of these two, which is \(1.04 + 0.68 = 1.72\)

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Key Concepts

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

Expected Values
When conducting a statistical analysis that involves comparing observed data to what we might expect theoretically, 'expected values' play a crucial role. Imagine trying to measure whether a coin is fair; you'd expect that it should land on heads about half of the time in a large number of flips.

In the context of the artery disease study on mummies, the expected values quantify how many cases of the disease we might predict to find if the ancient population had the same disease rate as today's population. The expected number of mummies with artery disease (6.4) is calculated by taking the total number of mummies and multiplying by the modern prevalence of coronary artery disease (40%). Similarly, the expected number of mummies without the disease (9.6) comes from the remainder who would not have been affected.

These calculations allow researchers to set a benchmark for comparison. Deviations from these expected numbers help pinpoint whether the rates of disease in the studied group are unusual, given the rates we observe today.
Coronary Artery Disease
Coronary artery disease (CAD) is one of the leading causes of mortality worldwide, typified by the buildup of plaque in the arteries that supply blood to the heart. This can lead to heart attacks, strokes, and other severe cardiac conditions.

Historically, many have attributed this disease to modern lifestyle factors like a diet high in processed foods and sedentary behavior. However, the discovery of hardened arteries in ancient mummies challenges this perspective, suggesting that CAD might have been present among humans long before these modern factors existed.

The study of these mummies provides valuable insights not only into the health and diet of ancient civilizations but also into the fundamental nature of coronary artery disease itself. It opens the door to questions about the genetic predisposition and environmental factors contributing to CAD.
Statistical Analysis

Understanding Chi-Square

Chi-square tests are statistical methods used to compare observed results with expected results. In the exercise with the mummies, the chi-square statistic provides a measure of how much the observed artery disease rates deviate from what was expected based on modern-day statistics.

By calculating the chi-square value (\(1.72\) in the mummy study), statisticians can assess the likelihood that any observed difference between the expected and the actual data is due to chance. A higher chi-square value typically indicates a greater deviation from the expected values, pointing towards a more significant result which may suggest a need to reject a null hypothesis鈥攊n this case, that the proportion of artery disease in mummies aligns with the modern rate.

It is a fundamental tool in epidemiological research and other fields that require the validation of hypotheses through observation and experiment, allowing conclusions to be drawn from empirical data.

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Most popular questions from this chapter

Suppose there is a theory that \(90 \%\) of the people in the United States dream in color. You survey a random sample of 200 people; 198 report that they dream in color, and 2 report that they do not. You wish to verify the claim made in the theory.

In the study referenced in exercise \(10.33\), researchers also collected data on use of apps to monitor diet and calorie intake. The data are reported in the table. Test the hypothesis that diet app use and gender are associated. Use a \(0.05\) significance level. $$ \begin{array}{ccc} \text { Use } & \text { Male } & \text { Female } \\ \hline \text { Yes } & 43 & 241 \\ \hline \text { No } & 50 & 84 \\ \hline \end{array} $$

Stradivarius violins, made in the 1700 s by a man of the same name, are worth millions of dollars. They are prized by music lovers for their uniquely rich, full sound. In September 2009 , an audience of experts took part in a blind test of violins, one of which was a Stradivarius. There were four other violins (modernday instruments) made of specially treated wood. When asked to pick the Stradivarius after listening to all five violins, 39 got it right and 113 got it wrong (Time, November 23,2009 ). a. If this group were just guessing, how many people (out of the 152 ) would be expected to guess correctly? And how many would be expected to guess incorrectly? b. Calculate the observed value of the chi-square statistic showing each step of the calculation.

In Montreal, Canada, an experiment was done with parents of children who were thought to have a high risk of committing crimes when they became teenagers (Tremblay et al., 1996 ). Some of the families were randomly assigned to receive parental training, and the others were not. Out of 43 children whose parents were randomly assigned to the parental training group, 6 had been arrested by the age of \(15 .\) Out of 123 children whose parents were not in the parental training group, 37 had been arrested by age 15 . a. Find and compare the percentages of children arrested by age \(15 .\) Is this what researchers might have hoped? b. Create a two-way table from the data, and test whether the treatment program is associated with arrests. Use a significance level of \(0.05\). c. Do a two-proportion \(z\) -test, testing whether the parental training lowers the rate of bad results. Use a significance level of \(0.05\). d. Explain the difference in the results of the chi-square test and the twoproportion \(z\) -test. e. Can you conclude that the treatment causes the better result? Why or why not?

An 2017 NPR/Marist poll asked a random sample of Americans if they had personally experienced sexual harassment in the workplace. The results are shown in the following table. $$ \begin{array}{lcc} & \begin{array}{c} \text { Personally Experienced Sexual Harassment } \\ \text { in the Workplace } \end{array} \\ \hline & \text { Yes } & \text { No } \\ \hline \text { Men } & 120 & 380 \\ \hline \text { Women } & 182 & 337 \\ \hline \end{array} $$ a. Find the percentage in each group who had personally experienced sexual harassment in the workplace. b. Test the hypothesis that experience of sexual harassment in the workplace is associated with gender at the \(0.05\) significance level.
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