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Chapter 5: Problem 61

The hypothetical data below are from a toxicity study designed to measure the effectiveness of different doses of a pesticide on mosquitoes. The table below summarizes the concentration of the pesticide, the sample sizes, and the number of critters dispatched. \begin{tabular}{lccccccc} \hline Concentration (g/cc) & \(0.10\) & \(0.15\) & \(0.20\) & \(0.30\) & \(0.50\) & \(0.70\) & \(0.95\) \\ Number of mosquitoes & 48 & 52 & 56 & 51 & 47 & 53 & 51 \\ Number killed & 10 & 13 & 25 & 31 & 39 & 51 & 49 \\ \hline \end{tabular} a. Make a scatterplot of the proportions of mosquitoes killed versus the pesticide concentration. b. Using the techniques introduced in this section, calculate \(y^{\prime}=\ln \left(\frac{p}{1-p}\right)\) for each of the concentrations and fit the line \(y^{\prime}=a+b\) (Concentration). What is the significance of a positive slope for this line? c. The point at which the dose kills \(50 \%\) of the pests is sometimes called LD50, for "Lethal dose \(50 \%\)." What would you estimate to be LD50 for this pesti-

Short Answer

Expert verified
The scatterplot illustrates a positive relationship between the concentration of pesticide and the proportion of mosquitoes killed. From the fitted line to the transformed values of \(y'\), a positive slope signifies an increase in the proportion killed as the concentration rises. The estimated LD50 is the concentration that would correspond to \(y' = ln \left( \frac{0.5}{1 - 0.5} \right)\) in the fitted model.

Step by step solution

01

Making the Scatterplot

With the data provided, create a scatterplot with concentration on the x-axis and proportion killed on the y-axis. The proportion is calculated by dividing the number killed by the number of mosquitoes for each concentration.
02

Calculating \(y'\) and Fitting the Line

Next, calculate \(y' = \ln \left( \frac{p}{1 - p} \right) \) for each concentration, where \( p \) is the proportion of mosquitoes killed. This transforms the proportions into a range that is easier to analyze. Then fit a line to the transformed data using the equation \(y' = a + b \times \text{Concentration} \). The slope of this line, \(b\), represents the relationship between the concentration of pesticide and the proportion of mosquitoes killed, with a positive slope indicating an increase in the proportion of mosquitoes killed with increased concentrations.
03

Estimating the LD50

Having calculated \( p \) as \( \frac{1}{1 + e^{-y'}}\), the LD50 value is the concentration that yields \( p = 0.5 \). To do this, find the concentration value that corresponds to \( y' = ln \left( \frac{0.5}{1 - 0.5} \right) \).

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Key Concepts

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

Scatterplot Creation
Scatterplots are essential tools in statistics for displaying the relationship between two variables. In the context of a toxicity study, creating a scatterplot can vividly illustrate how the proportion of mosquitoes killed varies with different concentrations of pesticide.

To create a scatterplot, each axis represents one of the variables: the x-axis displays the pesticide concentration, while the y-axis shows the proportion of mosquitoes killed. You calculate these proportions by dividing the number killed by the total number of mosquitoes at each concentration level. After plotting these points, you can visually assess the data distribution and any apparent trends, such as an increasing trend indicating more kills with higher concentrations.

Understanding the relationship between the two variables helps anticipate the outcomes at untested concentrations and is critical for identifying the effective dose range.
Logit Transformation
The logit transformation is a statistical technique used to handle proportions or probabilities. It transforms bounded data, lying between 0 and 1, to unbounded data which can span from negative to positive infinity. This transformation is given by the formula: \( y' = \text{ln} \bigg( \frac{p}{1 - p} \bigg) \), where \( p \) is the proportion of interest.

The transformation is particularly useful when dealing with data that exhibit non-linear relationships. By applying the logit transformation to the proportions of mosquitoes killed by different doses of pesticide, the data can be linearized, making it easier to fit a regression line and interpret the effects of different concentrations on mosquito mortality. A positive slope in this transformed scale indicates a stronger effect with increasing dose, which is crucial for understanding dose-response relationships in toxicology.
LD50 Estimation
LD50 stands for 'lethal dose, 50%', which is a statistic commonly used in toxicology to quantify the efficacy of a substance in killing pests. It refers to the dose required to kill half the members of a tested population. Estimating the LD50 is a vital step in understanding the potency of a pesticide.

After the logit transformation, estimating LD50 involves finding the dose corresponding to a 50% kill rate. This '50-50' point occurs where \( p = 0.5 \), so in the context of the logit transformation, where \( y' = \text{ln} \bigg( \frac{0.5}{1 - 0.5} \bigg) = 0 \) since logarithm of 1 is 0. By looking at the linear regression line fitted to the transformed data, one can determine the dose (concentration) that lines up with \( y' = 0 \). This concentration will provide a reasonable LD50 estimate, helping in the decision-making process regarding safe and effective usage levels of the pesticide.

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Most popular questions from this chapter

The following data on \(x=\) score on a measure of test anxiety and \(y=\) exam score for a sample of \(n=9\) students are consistent with summary quantities given in the paper "Effects of Humor on Test Anxiety and Performance"' (Psychological Reports [1999]: \(1203-1212\) ): $$ \begin{array}{llllrlllll} x & 23 & 14 & 14 & 0 & 17 & 20 & 20 & 15 & 21 \\ y & 43 & 59 & 48 & 77 & 50 & 52 & 46 & 51 & 51 \end{array} $$ Higher values for \(x\) indicate higher levels of anxiety. a. Construct a scatterplot, and comment on the features of the plot. b. Does there appear to be a linear relationship between the two variables? How would you characterize the relationship? c. Compute the value of the correlation coefficient. Is the value of \(r\) consistent with your answer to Part (b)? d. Is it reasonable to conclude that test anxiety caused poor exam performance? Explain.

The accompanying data are \(x=\) cost (cents per serving) and \(y=\) fiber content (grams per serving) for 18 high-fiber cereals rated by Consumer Reports (www .consumerreports.org/health). $$ \begin{array}{cc} \begin{array}{c} \text { Cost per } \\ \text { serving } \end{array} & \begin{array}{c} \text { Fiber per } \\ \text { serving } \end{array} \\ \hline 33 & 7 \\ 46 & 10 \\ 49 & 10 \\ 62 & 7 \\ 41 & 8 \\ 19 & 7 \\ 77 & 12 \\ 71 & 12 \\ 30 & 8 \\ 53 & 13 \\ 53 & 10 \\ 67 & 8 \\ 43 & 12 \\ 48 & 7 \\ 28 & 14 \\ 54 & 7 \\ 27 & 8 \\ 58 & 8 \\ \hline \end{array} $$ a. Compute and interpret the correlation coefficient for this data set. b. The serving size differed for the different cereals, with serving sizes varying from \(1 / 2\) cup to \(1^{1 / 4}\) cups. Converting price and fiber content to "per cup" rather than "per serving" results in the accompanying data. Is the correlation coefficient for the per cup data greater than or less than the correlation coefficient for the per serving data? $$ \begin{array}{cc} \text { Cost per Cup } & \text { Fiber per Cup } \\ \hline 9.3 & 44 \\ 10 & 46 \\ 10 & 49 \\ 7 & 62 \\ 6.4 & 32.8 \\ 7 & 19 \\ 12 & 77 \\ 9.6 & 56.8 \\ 8 & 30 \\ 13 & 53 \\ 10 & 53 \\ 8 & 67 \\ 12 & 43 \\ 7 & 48 \\ 28 & 56 \\ 7 & 54 \\ 16 & 54 \\ 10.7 & 77.3 \\ \hline \end{array} $$

The paper "Effects of Canine Parvovirus (CPV) on Gray Wolves in Minnesota" (journal of Wildlife Management [1995]: \(565-570\) ) summarized a regression of \(y=\) percentage of pups in a capture on \(x=\) percentage of CPV prevalence among adults and pups. The equation of the least-squares line, based on \(n=10\) observations, was \(\hat{y}=62.9476-0.54975 x\), with \(r^{2}=.57\). a. One observation was \((25,70)\). What is the corresponding residual? b. What is the value of the sample correlation coefficient? c. Suppose that \(\mathrm{SSTo}=2520.0\) (this value was not given in the paper). What is the value of \(s_{e}\) ?

\(5.19\) - The accompanying data on \(x=\) head circumference \(z\) score (a comparison score with peers of the same age - a positive score suggests a larger size than for peers) at age 6 to 14 months and \(y=\) volume of cerebral grey matter (in \(\mathrm{ml}\) ) at age 2 to 5 years were read from a graph in the article described in the chapter introduction (journal of the American Medical Association [2003]). \begin{tabular}{lc} Cerebral Grey Matter (ml) \(2-5\) yr & Head Circumference \(z\) Scores at \(6-14\) Months \\ \hline 680 & \(-.75\) \\ 690 & \(1.2\) \\ 700 & \(-.3\) \\ 720 & \(.25\) \\ 740 & \(.3\) \\ 740 & \(1.5\) \\ 750 & \(1.1\) \\ 750 & \(2.0\) \\ 760 & \(1.1\) \\ 780 & \(1.1\) \\ 790 & \(2.0\) \\ 810 & \(2.1\) \\ 815 & \(2.8\) \\ 820 & \(2.2\) \\ 825 & \(.9\) \\ 835 & \(2.35\) \\ 840 & \(2.3\) \\ 845 & \(2.2\) \\ \hline \end{tabular} a. Construct a scatterplot for these data. b. What is the value of the correlation coefficient? c. Find the equation of the least-squares line. d. Predict the volume of cerebral grey matter at age 2 to 5 years for a child whose head circumference \(z\) score at age 12 months was \(1.8\). e. Explain why it would not be a good idea to use the least-squares line to predict the volume of grey matter for a child whose head circumference \(z\) score was \(3.0\).

Anabolic steroid abuse has been increasing despite increased press reports of adverse medical and psychiatric consequences. In a recent study, medical researchers studied the potential for addiction to testosterone in hamsters (Neuroscience \([2004]: 971-981)\). Hamsters were allowed to self-administer testosterone over a period of days, resulting in the death of some of the animals. The data below show the proportion of hamsters surviving versus the peak self-administration of testosterone \((\mu \mathrm{g}) .\) Fit a logistic regression equation and use the equation to predict the probability of survival for a hamster with a peak intake of \(40 \mu \mathrm{g}\). \begin{tabular}{cccc} \multicolumn{4}{c} { Survival } \\ Peak Intake (micrograms) & Proportion \((p)\) & \(\frac{p}{1-p}\) & \(y^{\prime}=\ln \left(\frac{p}{1-p}\right)\) \\ \hline 10 & \(0.980\) & \(49.0000\) & \(3.8918\) \\ 30 & \(0.900\) & \(9.0000\) & \(2.1972\) \\ 50 & \(0.880\) & \(7.3333\) & \(1.9924\) \\ 70 & \(0.500\) & \(1.0000\) & \(0.0000\) \\ 90 & \(0.170\) & \(0.2048\) & \(-1.5856\) \\ \hline \end{tabular}
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