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Chapter 13: Problem 37

The results of an experiment to assess the effect of crude oil on fish parasites are described in the article "Effects of Crude Oils on the Gastrointestinal Parasites of Two Species of Marine Fish" (J. Wildlife Diseases, 1983: 253-258). Three treatments (corresponding to populations in the procedure described) were compared: (1) no contamination, (2) contamination by 1-year-old weathered oil, and (3) contamination by new oil. For each treatment condition, a sample of fish was taken, and then each fish was classified as either parasitized or not parasitized. Data compatible with that in the article is given. Does the data indicate that the three treatments differ with respect to the true proportion of parasitized and nonparasitized fish? Test using \(\alpha=.01\). $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Parasitized } & \text { Nonparasitized } \\\ \hline \text { Control } & 30 & 3 \\ \text { Old oil } & 16 & 8 \\ \text { New oil } & 16 & 16 \end{array} $$

Short Answer

Expert verified
Yes, the treatments differ. Chi-Square test shows significant differences among parasitized proportions.

Step by step solution

01

State the Hypotheses

We need to determine if there's a difference between proportions of parasitized fish in different treatments. The null hypothesis (H_0) is that the proportions of parasitized fish are the same across all treatments. The alternative hypothesis (H_a) is that at least one treatment has a different proportion.\[ H_0: p_1 = p_2 = p_3 \]\[ H_a: ext{at least one } p_i ext{ is different} \]
02

Compute Proportions for Each Treatment

Calculate the proportion of parasitized fish for each treatment.- Control Group: \( p_1 = \frac{30}{30+3} \approx 0.91 \)- Old Oil Group: \( p_2 = \frac{16}{16+8} \approx 0.67 \)- New Oil Group: \( p_3 = \frac{16}{16+16} = 0.50 \)
03

Use Chi-Square Test for Homogeneity

Since we want to compare proportions across multiple groups, we use a Chi-Square test for homogeneity. Compute the expected counts assuming the null hypothesis is true, and calculate the Chi-Square statistic:\[ \text{Total observed parasitized} = 30+16+16 = 62 \\text{Total observed nonparasitized} = 3+8+16 = 27 \\text{Overall proportion of parasitized} = \frac{62}{62+27} \approx 0.696 \]Calculate expected counts for each cell based on overall proportion.
04

Calculate Expected Counts

Calculate expected counts under the null hypothesis for each cell in the table, and then use these to compute the Chi-Square statistic:- Control Expected: Parasitized: \((33)(0.696) \approx 22.968\), Nonparasitized: \((33)(0.304) \approx 10.032\)- Old Oil Expected: Parasitized: \((24)(0.696) \approx 16.704\), Nonparasitized: \((24)(0.304) \approx 7.296\)- New Oil Expected: Parasitized: \((32)(0.696) = 22.272\), Nonparasitized: \((32)(0.304) = 9.728\)
05

Compute Chi-Square Statistic

Calculate the Chi-Square statistic using the observed and expected counts:\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Substitute the respective observed (O) and expected (E) counts into the formula. Compute for each treatment and total the values to get the Chi-Square statistic.
06

Draw Conclusion

Determine the critical value from the Chi-Square distribution table with \( df = 2 \) (degrees of freedom = number of groups - 1) at \( \alpha = 0.01 \). Compare the calculated Chi-Square statistic to the critical value:- If \( \chi^2_{calculated} > \chi^2_{critical} \), reject the null hypothesis.- If \( \chi^2_{calculated} \leq \chi^2_{critical} \), fail to reject the null hypothesis.

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Key Concepts

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

Chi-Square Test
A Chi-Square test is a statistical method used to examine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories of a contingency table. In the context of our exercise, we want to test if different treatments of crude oil result in different proportions of parasitized fish.
First, we compute the observed counts of parasitized and nonparasitized fish for each treatment: no contamination, old oil, and new oil. Next, we calculate the expected counts under the assumption that the null hypothesis is true, which states that all treatments have the same proportion of parasitized fish.
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 count and \(E_i\) represents the expected count. By comparing the calculated Chi-Square value to a critical value from the Chi-Square distribution table, we can determine whether to accept or reject the null hypothesis. If the calculated Chi-Square statistic is greater than the critical value, it indicates a significant difference between the treatments.
Statistical Significance
Statistical significance is a mathematical measure that provides insights into whether the difference in data outcomes is due to a specific cause or merely by chance. In the crude oil and fish parasites study, we assess statistical significance by testing if the treatment methods significantly affect the proportion of parasitized fish.
We utilize a significance level (\(\alpha\)), which is a threshold probability for rejecting the null hypothesis. In this case, \(\alpha = 0.01\) is chosen, indicating that there is a 1% risk of concluding that a difference exists when there is none.
To test for significance, we calculate the Chi-Square statistic and compare it to the critical value associated with our chosen \(\alpha\). If the calculated \(\chi^2\) exceeds this critical value, the result is considered statistically significant, leading us to reject the null hypothesis. This means that at least one treatment effect differs significantly from the others, providing strong evidence that treatment impacts parasitism.
Treatment Effect Analysis
Treatment effect analysis examines how different treatments influence outcomes, focusing on the measured difference between conditions. In this study on fish parasites, three treatments are analyzed: no oil contamination, old oil contamination, and new oil contamination.
To understand the treatment effects, the proportions of parasitized fish are calculated for each treatment group. These proportions give an initial view of how each oil treatment affects the fish compared to a control group.
We then apply the Chi-Square test to statistically verify whether these observed differences are significant. By clarifying which treatments have a significant impact, we can deduce the effectiveness of each oil treatment on fish health. It's important because this information can guide decisions on environmental management, determining the less detrimental treatment for marine ecosystems.

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Most popular questions from this chapter

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 NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002 , the top-ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3 . Let \(P_{i j}\) denote the probability that the team ranked \(i\) in its region is victorious in its game against the team ranked \(j\). Once the \(P_{i j}\) 's are available, it is possible to compute the probability that any particular seed wins its regional tournament (a complicated calculation because the number of outcomes in the sample space is quite large). The paper "Probability Models for the NCAA Regional Basketball Tournaments"(Amer. Statist., 1991: 35-38) proposed several different models for the \(P_{i j}\) 's. a. One model postulated \(P_{i j}=.5-\lambda(i-j)\) with \(\lambda=\frac{1}{32}\) (from which \(P_{16,1}=\frac{1}{32}, P_{16,2}=\frac{2}{32}\), etc.). Based on this, \(P\) (seed #1 wins) \(=.27477\), \(P(\) seed \(\\# 2\) wins \()=.20834\), and \(P\) (seed #3 wins \()=.15429\). Does this model appear to provide a good fit to the data? b. A more sophisticated model has \(P_{i j}=.5+\) \(.2813625\left(z_{i}-z_{j}\right)\), where the \(z\) 's are measures of relative strengths related to standard normal percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the range of probabilities is the same as for the model in part (a)]. The resulting probabilities of seeds 1,2 , or 3 winning their regional tournaments are \(.45883, .18813\), and \(.11032\), respectively. Assess the fit of this model.

The article "Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients' Birthdays"' (Psych. Rep., 1992: 944鈥�946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient's birthday. Assuming a 365day year (i.e., excluding leap year), in the absence of any relation, a patient's admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of \(11,24,69\), and 96 for categories 1,2 , 3 , and 4, respectively. State and test the relevant hypotheses using a significance level of \(.01\).

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.

Suppose that in a particular state consisting of four distinct regions, a random sample of \(n_{k}\) voters is obtained from the \(k\) th region for \(k=1,2,3,4\). Each voter is then classified according to which candidate \((1,2\), or 3\()\) he or she prefers and according to voter registration ( \(1=\) Dem., \(2=\) Rep., \(3=\) Indep.). Let \(p_{i j k}\) denote the proportion of voters in region \(k\) who belong in candidate category \(i\) and registration category \(j\). The null hypothesis of homogeneous regions is \(H_{0}\) : \(p_{i j 1}=p_{i j 2}=p_{i j 3}=p_{i j 4}\) for all \(i, j\) (i.e., the proportion within each candidate/registration combination is the same for all four regions). Assuming that \(H_{0}\) is true, determine \(\hat{p}_{i j k}\) and \(\hat{e}_{i j k}\) as functions of the observed \(n_{i j k}\) 's, and use the general rule of thumb to obtain the number of degrees of freedom for the chi-squared test.
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