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Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. What Is the Effect of Including Some Indifferent Fish? In the experiment described above under Fish Democracies, the schools of fish in the study with an opinionated minority and a less passionate majority picked the majority option only about \(17 \%\) of the time. However, when groups also included 10 fish with no opinion, the schools of fish picked the majority option \(61 \%\) of the time. We want to estimate the effect of adding the fish with no opinion to the group, which means we want to estimate the difference in the two proportions. We learn from the study that the standard error for estimating this difference is about \(0.14 .\) Define the parameter we are estimating, give the best point estimate, and find and interpret a \(95 \%\) confidence interval. Is it plausible that adding indifferent fish really has no effect on the outcome?

Short Answer

Expert verified
The best point estimate for the effect of adding the fish with no opinion to the group is 0.44. A 95% confidence interval for this estimate is (0.1656, 0.7144). Since this interval does not include zero, we conclude it is plausible that adding indifferent fish truly affects the outcome and the majority choice prevails more often.

Step by step solution

01

Define the Parameter

The parameter in this case is the difference in the proportions of time the school of fish opted for the majority option. This can be represented as \( \Delta p = p_2 - p_1 \), where \( p_1 \) is the proportion in the absence of indifferent fish (17% or 0.17) and \( p_2 \) is the proportion in the presence of indifferent fish (61% or 0.61).
02

Compute the Point Estimate

The point estimate is the observed value of the parameter and it can be calculated by substituting the provided values into the parameter equation: \( \Delta p = p_2 - p_1 = 0.61 - 0.17 = 0.44\). The point estimate is 0.44.
03

Compute the Confidence Interval

A 95% confidence interval can be computed using the formula for the difference in proportions: \( CI = \Delta p \pm Z * SE \), where \( Z \) is the Z-score from the Z-table corresponding to the desired confidence level (1.96 for 95% confidence), and \( SE \) is the given standard error (0.14). Plugging these values into the formula gives us: \( CI = \Delta p \pm Z * SE = 0.44 \pm 1.96 * 0.14\) which computes to \( CI = 0.44 \pm 0.2744 \). Hence the 95% confidence interval is (0.1656, 0.7144).
04

Interpreting the Interval and Conclusion

The 95% confidence interval (0.1656, 0.7144) suggests that if we were to conduct the same experiment 100 times, we would expect the difference in proportions to fall within this range 95 times. The interval does not include 0, implying we can be 95% confident that adding indifferent fish has an impact on the decision making of the group. It is therefore plausible that adding indifferent fish does affect the outcome.

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Key Concepts

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

Statistical Significance
Understanding statistical significance is crucial when interpreting the results of experiments, including studies like the one involving golden shiners. In this context, statistical significance is used to determine whether the observed difference in outcomes—like the shift in the fish's group decision due to the presence of indifferent fish—is unlikely to have occurred by chance. A result is deemed statistically significant if the probability of the observed effect happening due to random variation is low.

In our fish democracy study, the difference in proportions showed that the group more frequently chose the majority option when indifferent fish were added. The confidence interval computed did not include 0, which suggests that this result is unlikely to be due to random chance. Therefore, we can infer that the presence of indifferent fish significantly influences group decisions. When evaluating this significance, researchers typically set a threshold, or alpha level (commonly 0.05), to determine the cut-off for what they would call statistically significant.
Difference in Proportions
The difference in proportions is a measure used to compare the rates of certain outcomes between two groups. In the fish study, this concept is used to compare the frequencies at which two different groups of fish choose the majority option. Specifically, the parameter of interest, denoted as \( \Delta p \), is the difference between the proportion of times the group with indifferent fish chose the majority sign (61%) and the times the group without them did so (17%).

The calculated difference, 0.44, represents a considerable shift in behavior and enables researchers to assess the magnitude of the effect that indifferent fish have on the group's decisions. It is important to articulate the meaning of each proportion to ensure clarity when assessing the practical implications of such a difference, especially in educational materials aimed at helping students understand these comparisons.
Standard Error
The standard error is a statistic that measures the variability or uncertainty around an estimate. It is essential when constructing a confidence interval as it provides a sense of how precise our estimate is likely to be. In the experiment with the fish, the standard error is given as 0.14 for the difference in proportions. This figure is integral to determining the reliability of our difference measurement; a smaller standard error means we can be more confident that our estimated difference in proportions is close to the true difference in the population.

For students grappling with this concept, understanding that the standard error serves as the yardstick for gauging variability—one of the key aspects of statistical estimation—can be quite enlightening. Moreover, the relationship between sample size and standard error is inverse; increasing the sample size generally results in a lower standard error, implying a more precise estimate.
Parameter Estimation
Parameter estimation in statistics involves using sample data to approximate the values of population parameters. In this fish study, the parameter we want to estimate is the true difference in the proportion of times the fish schools pick the majority decision under two scenarios: with and without the presence of indifferent fish. The process began with defining the parameter, resulting in our point estimate of 0.44, followed by constructing a 95% confidence interval around this point estimate.

Interpreting the interval is just as important as computation: the range from 0.1656 to 0.7144 tells us that the true difference in proportions likely falls within these values with a high degree of confidence. The fact that the interval does not include zero underscores the impact uncommitted members have on the decision-making process. For students studying statistics, grasping the methods and importance of parameter estimation is fundamental to understanding how we draw conclusions about entire populations based on sample observations.

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Most popular questions from this chapter

Exercises 3.112 to 3.115 give information about the proportion of a sample that agree with a certain statement. Use StatKey or other technology to find a confidence interval at the given confidence level for the proportion of the population to agree, using percentiles from a bootstrap distribution. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. Find a \(90 \%\) confidence interval if 112 agree and 288 disagree in a random sample of 400 people.

College Graduates In Example 3.1 on page 197, we see that \(27.5 \%\) of US adults are college graduates. (a) Use StatKey or other technology to generate a sampling distribution for the sample proportion of college graduates using a sample size of \(n=50 .\) Generate at least 1000 sample proportions. Give the shape and center of the sampling distribution and give the standard error. (b) Repeat part (a) using a sample size of \(n=500\).

Is a Car a Necessity? A random sample of \(n=1483\) adults in the US were asked whether they consider a car a necessity or a luxury, \({ }^{31}\) and we find that a \(95 \%\) confidence interval for the proportion saying that it is a necessity is 0.83 to \(0.89 .\) Explain the meaning of this confidence interval in the appropriate context.

\(\mathbf{3 . 1 2 3}\) What Proportion Have Pesticides Detected? In addition to the quantitative variable pesticide concentration, the researchers also report whether or not the pesticide was detected in the urine (at standard detection levels). Before the participants started eating organic, 111 of the 240 measurements (combining all pesticides and people) yielded a positive pesticide detection. While eating organic, only 24 of the 240 measurements resulted in a positive pesticide detection. (a) Calculate the sample difference in proportions: proportion of measurements resulting in pesticide detection while eating non- organic minus proportion of measurements resulting in pesticide detection while eating organic. (b) Figure 3.33 gives a bootstrap distribution for the difference in proportions, based on \(1000 \mathrm{sim}-\) ulated bootstrap samples. Approximate a \(98 \%\) confidence interval. (c) Interpret this interval in context.

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. $$ \bar{x}=55 \text { and the standard error is } 1.5 . $$

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