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Pesticides The following table, on the presence of pesticide residues in samples of organic and conventional food, was analyzed in Examples 2 and 3 in Chapter 3 . \begin{tabular}{lrr} \hline & \multicolumn{2}{c} { Pesticides } \\ \cline { 2 - 3 } Food Type & Yes & \multicolumn{1}{c} { No } \\ \hline Organic & 29 & 98 \\ Conventional & 19485 & 7086 \\ \hline \end{tabular} a. Compute the relative risk and interpret. b. Find the reciprocal of your answer to part a and interpret. c. Compute the odds ratio and interpret. d. Find the reciprocal of your answer to part \(\mathrm{c}\) and interpret. e. What's wrong with this statement: The proportion of conventional food samples with pesticide residues present is more than 9 times larger than the proportion of organic food samples with pesticide residues present.

Short Answer

Expert verified
a: RR ≈ 3.216. b: Reciprocal RR ≈ 0.311. c: OR ≈ 9.283. d: Reciprocal OR ≈ 0.108. e: The statement misuses proportions instead of odds ratio, true RR is ~3.216.

Step by step solution

01

Calculate Proportions

First, calculate the proportion of samples with pesticide residues in both organic and conventional foods.\[\text{Proportion}_{\text{Organic (Yes)}} = \frac{29}{29+98} = \frac{29}{127} \approx 0.228\] \[\text{Proportion}_{\text{Conventional (Yes)}} = \frac{19485}{19485+7086} = \frac{19485}{26571} \approx 0.733\] These calculations will be used for further analysis.
02

Compute Relative Risk

Now, calculate the relative risk (RR) which is the risk of pesticides in conventional foods versus organic foods. The formula is: \[\text{RR} = \frac{\text{Proportion}_{\text{Conventional (Yes)}}}{\text{Proportion}_{\text{Organic (Yes)}}} = \frac{0.733}{0.228} \approx 3.216\] This implies conventional foods are approximately 3.216 times more likely to contain pesticide residues compared to organic foods.
03

Reciprocal of Relative Risk

The reciprocal of the relative risk (RR) tells us the risk of pesticides in organic foods relative to conventional foods.\[\text{Reciprocal of RR} = \frac{1}{3.216} \approx 0.311\] This implies organic foods have about 31.1% the risk of having pesticide residues compared to conventional foods.
04

Compute Odds Ratio

Calculate the odds of pesticide residues in both types of food and then the odds ratio (OR). \[\text{Odds}_{\text{Organic}} = \frac{29}{98} = 0.296\] \[\text{Odds}_{\text{Conventional}} = \frac{19485}{7086} \approx 2.749\] \[\text{Odds Ratio (OR)} = \frac{\text{Odds}_{\text{Conventional}}}{\text{Odds}_{\text{Organic}}} = \frac{2.749}{0.296} \approx 9.283\] This indicates that the odds of pesticide residues are approximately 9.283 times higher in conventional foods than in organic foods.
05

Reciprocal of Odds Ratio

Find the reciprocal of the odds ratio to compare the odds for organic foods relative to conventional foods. \[\text{Reciprocal of OR} = \frac{1}{9.283} \approx 0.108\] This suggests that the odds of pesticide residues in organic food is about 10.8% that of conventional food.
06

Evaluate Statement

Examine the statement: "The proportion of conventional food samples with pesticide residues present is more than 9 times larger than the proportion of organic food samples with pesticide residues present." The statement incorrectly uses proportions to represent what is more appropriately expressed through the odds ratio. The relative risk of 3.216, not 9, applies to proportions.

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Key Concepts

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

Relative Risk
When we talk about relative risk in statistical analysis, we're assessing the likelihood of an event happening in one group relative to another. Here, we compare the probability of finding pesticide residues in conventional food to that in organic food.
To calculate relative risk (RR), we divide the probability (or proportion) of the event in our group of interest by the probability of the event in the comparison group. In this exercise, the relative risk was calculated as follows:
  • Proportion of conventional foods with pesticides: 0.733
  • Proportion of organic foods with pesticides: 0.228
Thus, the RR is: \[ RR = \frac{0.733}{0.228} \approx 3.216 \] This means that conventional foods are about 3.216 times more likely to contain pesticide residues compared to organic foods. This simple metric helps us quickly grasp and compare the risk across these two categories. It’s an essential tool in epidemiology and public health assessments.
Odds Ratio
An odds ratio (OR) is another extremely valuable metric in statistical analysis. It compares the odds of an event occurring in one group to the odds of it occurring in another. This is particularly useful in case-control studies.
In our current scenario, we want to compare the odds of pesticide residues in conventional versus organic foods. To find the odds for each:
  • The odds of pesticides in organic foods: \(\frac{29}{98} = 0.296\)
  • The odds of pesticides in conventional foods: \(\frac{19485}{7086} \approx 2.749\)
The odds ratio is calculated by dividing these odds: \[ OR = \frac{2.749}{0.296} \approx 9.283 \] This indicates the odds of finding pesticide residues are approximately 9.283 times higher in conventional foods than in organic foods. Or, put differently, if you select a conventional food sample, the chance it contains pesticide residues is notably greater compared to an organic sample. The odds ratio provides deep insights into the strength of the association between the food type and pesticide presence.
Proportions
Proportions give us a fundamental understanding of how often an event occurs in a particular group, expressed as a fraction or percentage. In many situations, understanding proportions can help make sense of large datasets in a simplified and quantitative way.
In the context of the pesticide residue data:
  • Proportion of organic samples with pesticides: \(\frac{29}{127} \approx 0.228\)
  • Proportion of conventional samples with pesticides: \(\frac{19485}{26571} \approx 0.733\)
This means that approximately 22.8% of organic food samples tested contained pesticides, while about 73.3% of conventional samples contained pesticides.
Using proportions is key when interpreting data in epidemiological studies. They allow for a straightforward comparison of the prevalence of events between different groups. However, it is crucial to use these figures correctly within their context. In our exercise, the phrase "The proportion of conventional food samples with pesticide residues present is more than 9 times larger than the proportion of organic food samples with pesticide residues present", is misleading. Such a statement incorrectly applies what is true for the odds ratio (in terms of comparing odds, not proportions), highlighting the importance of distinguishing between probability and odds in statistical communication.

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Most popular questions from this chapter

In an experiment on chlorophyll inheritance in corn, for 1,103 seedlings of self-fertilized heterozygous green plants, 854 seedlings were green and 249 were yellow. Theory predicts that \(75 \%\) of the seedlings would be green. a. Specify a null hypothesis for testing the theory. b. Find the value of the chi-squared goodness-of-fit statistic and report its \(d f\). c. Report the P-value and interpret.

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}

Study hours and grades The following table shows data on study hours per week and the effect on grades, with expected cell counts given underneath the observed counts for 200 college students in a study conducted by Washington's Public Interest Research Group (PIRG). \begin{tabular}{lcllc} \hline & \multicolumn{3}{c} { Effect on grades } & \\ \cline { 2 - 4 } Study hours per week & Positive & None & Negative & Total \\ \hline \(1-15\) & 26 & 50 & 14 & 90 \\ & 23.9 & 43.2 & 23.0 & \\ \(16-24\) & 16 & 27 & 17 & 60 \\ & 15.9 & 28.8 & 15.3 & \\ \(25-34\) & 11 & 19 & 20 & 50 \\ & 13.3 & 24.0 & 12.8 & \\ Total & 53 & 96 & 51 & 200 \\ \hline \end{tabular} 2002 ) (Source: USA Today, April 17 , a. Suppose the variables were independent. Explain what this means in this context. b. Explain what is meant by an expected cell count. Show how to get the expected cell count for the first cell, for which the observed count is 26 . c. Compare the expected cell frequencies to the observed counts. Based on this, what is the profile of subjects who tend to have (i) positive effect on grades than independence predicts and (ii) negative effect on grades than independence predicts.

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True or false: \(X^{2}=0\) The null hypothesis for the test of independence between two categorical variables is \(\mathrm{H}_{0}: X^{2}=0\) for the sample chi-squared statistic \(X^{2}\). (Hint: Do hypotheses refer to a sample or the population?)

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