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Chapter 6: Problem 151

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

Short Answer

Expert verified
The difference in proportions for the samples A and B is 0.15. The standard error of this difference computed using the formula is approximately 0.061. The standard error computed using a bootstrap distribution should be approximately the same, reaffirming the accuracy of the formula.

Step by step solution

01

Calculate the proportions

First, calculate the proportions for both samples A and B. The proportion is given by the number of successes divided by the total number of trials. For sample A, it's \(90/120=0.75\). For sample B, it's \(180/300 = 0.60\). The difference in proportions is \(0.75 - 0.60 = 0.15\).
02

Compute the standard error using the formula

The formula for the standard error of a difference in proportions is \(\sqrt{((p_1(1-p_1))/n_1) + ((p_2(1-p_2))/n_2)}\), where \(p_1\) and \(p_2\) are the proportions of the two samples, and \(n_1\) and \(n_2\) are the sizes of the two samples. Plugging in the calculated proportions and sample sizes gives: \(\sqrt{((0.75)(1-0.75))/120 + ((0.60)(1-0.60))/300} \approx 0.061\)
03

Generate a bootstrap distribution

Using software like StatKey, generate a bootstrap distribution from the observed samples A and B, calculate the difference in proportions for each resample, and compute the standard error of these differences. This simulation-based standard error is likely to be quite close to the standard error computed in step 2, reaffirming the accuracy of the standard error formula.

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Key Concepts

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

Sample Proportions
Sample proportions are a fundamental concept in statistics, capturing the ratio of successes to the total number of trials in a given sample. Let鈥檚 dive deeper into what they mean and how to calculate them.
  • To compute a sample proportion, divide the number of 'successes' by the total sample size (n). For instance, in Sample A with 90 successes out of 120 trials, the sample proportion is calculated as \( \frac{90}{120} = 0.75 \).
  • Similarly, for Sample B, with 180 successes out of 300 trials, the proportion is \( \frac{180}{300} = 0.60 \).
These sample proportions provide essential information about each sample. They form the basis for comparing different samples, particularly when analyzing differences in proportions from two samples. This insight can assist in making informed predictions or testing statistical hypotheses. Remember that the higher the number of trials, the more reliable the sample proportion is likely to be as a representation of the underlying population proportion.
Standard Error
The standard error plays a crucial role in statistical inference, offering a measure of the variability or dispersion around a statistic, like a sample proportion. Here鈥檚 a closer look:
  • The standard error of a sample proportion represents how much the sample proportion is expected to vary from the actual population proportion. It's akin to the standard deviation, but instead, it applies to a statistic rather than raw data.
  • For differences in proportions, the formula to calculate the standard error is \[ \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \] where \( p_1 \) and \( p_2 \) are the sample proportions, and \( n_1 \) and \( n_2 \) are the sample sizes.
  • Using our examples: \( p_1 = 0.75, n_1 = 120, p_2 = 0.60, n_2 = 300 \), the standard error is approximately 0.061.
The standard error provides an essential check on our difference in sample proportions, gauging the reliability of our estimate. A smaller standard error indicates that our sample estimate is more precise, while a larger one suggests greater uncertainty.
Difference in Proportions
When comparing two samples, the difference in proportions becomes an invaluable statistic. It helps us to quantify the extent to which two different populations behave similarly or differently under a given measure. Here鈥檚 how to unravel this concept:
  • The difference in proportions between two samples, such as A and B, is calculated by subtracting the second sample proportion from the first. In our case, \( 0.75 - 0.60 = 0.15 \).
  • This difference indicates how much more likely it is for an event to occur in one group compared to another. A positive value means it is more likely in the first sample; a negative one would mean the opposite.
  • Performing a bootstrap distribution can enhance understanding by simulating new sample differences. Software tools, like StatKey, allow us to create a distribution of these simulated differences to visually and statistically verify our calculated standard error.
Understanding differences in proportions allows researchers to draw meaningful comparisons and conclusions about their sampled populations. Thanks to methods like bootstrapping, we can double-check our analytical results, reinforcing their accuracy and reliability.

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Most popular questions from this chapter

Can Malaria Parasites Control Mosquito Behavior? Are malaria parasites able to control mosquito behavior to their advantage? A study \(^{43}\) investigated this question by taking mosquitos and giving them the opportunity to have their first "blood meal" from a mouse. The mosquitoes were randomized to either eat from a mouse infected with malaria or an uninfected mouse. At several time points after this, mosquitoes were put into a cage with a human and it was recorded whether or not each mosquito approached the human (presumably to bite, although mosquitoes were caught before biting). Once infected, the malaria parasites in the mosquitoes go through two stages: the Oocyst stage in which the mosquito has been infected but is not yet infectious to others and then the Sporozoite stage in which the mosquito is infectious to others. Malaria parasites would benefit if mosquitoes sought risky blood meals (such as biting a human) less often in the Oocyst stage (because mosquitos are often killed while attempting a blood meal) and more often in the Sporozoite stage after becoming infectious (because this is one of the primary ways in which malaria is transmitted). Does exposing mosquitoes to malaria actually impact their behavior in this way? (a) In the Oocyst stage (after eating from mouse but before becoming infectious), 20 out of 113 mosquitoes in the group exposed to malaria approached the human and 36 out of 117 mosquitoes in the group not exposed to malaria approached the human. Calculate the Z-statistic. (b) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Oocyst stage approaching the human is lower in the group exposed to malaria. (c) In the Sporozoite stage (after becoming infectious), 37 out of 149 mosquitoes in the group exposed to malaria approached the human and 14 out of 144 mosquitoes in the group not exposed to malaria approached the human. Calculate the z-statistic. (d) Calculate the p-value for testing whether this provides evidence that the proportion of mosquitoes in the Sporozoite stage approaching the human is higher in the group exposed to malaria. (e) Based on your p-values, make conclusions about what you have learned about mosquito behavior, stage of infection, and exposure to malaria or not. (f) Can we conclude that being exposed to malaria (as opposed to not being exposed to malaria) causes these behavior changes in mosquitoes? Why or why not?

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