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Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

Short Answer

Expert verified
The short answer will depend on the actual data obtained in Step 1. Therefore, it can't be provided without the bootstrap distribution. However, the process will provide an understanding of whether the theoretical assumptions of standard error made by the Central Limit Theorem hold true for the given data.

Step by step solution

01

Calculate Bootstrap Standard Error

First, generate a bootstrap distribution of sample proportions. This process will be done by resampling the given data numerous times, and each time calculating the sample proportion. Collect all these new proportions in order to form the bootstrap distribution. Once done, calculate the standard deviation of this bootstrap distribution, which serves as the estimated standard error.
02

Calculate Theoretical Standard Error

Use the Central Limit Theorem (CLT) to calculate the theoretical standard error. According to the CLT, the standard error is calculated as \(SE = \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}\), where \(n\) is the sample size and \(\hat{p}\) is the sample proportion. Substituting \(n = 120\) and \(\hat{p} = 0.583\) into the formula, solve for \(SE\).
03

Compare Standard Errors

Compare the bootstrap standard error from Step 1 with the theoretical standard error from Step 2. The comparison will give an insight into how well the results from the actual data align with the theoretical model.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that describes how the distribution of sample means becomes increasingly normal as the sample size grows, regardless of the population's distribution. When students deal with real-world data, the CLT allows them to make inferences about population parameters using sample statistics.

Consider an example where a student is interested in the proportion of home team wins in soccer. If they take multiple samples, the means of those samples tend to form a normal distribution according to the CLT. This is especially helpful because normal distributions have well-understood properties, allowing us to make predictions and construct confidence intervals. For a large enough sample size, the sample mean will approximate a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. The formula given by the CLT to calculate standard error of the sample proportion is an essential tool for students to understand the variability of their estimates.
Standard Error
The standard error (SE) of a statistic is a measure of the variability or spread of that statistic if we were to take many samples from a population. It is crucial for interpreting the accuracy of an estimate from a single sample. In the context of our soccer wins example, the standard error gives an indication of how much the sample proportion of home team wins (hat{p}) might vary from sample to sample.

To calculate the standard error according to the Central Limit Theorem, one can use the formula SE = sqrt{frac{hat{p}(1 - hat{p})}{n}}, where n is the sample size. This formula takes into account the observed sample proportion and the size of the sample to provide an estimate of the standard error. Understanding this concept is essential since the SE is often used in constructing the confidence intervals and it gives a sense of how precise our estimate of the population parameter is.
Sample Proportion
Sample proportion (hat{p}) is a statistic that estimates the true proportion (p) of a certain characteristic within a population based just on a sample. In research studies or exercises like the example given, using the sample proportion is practical and common because it's not always feasible to gather data from an entire population. For instance, a sample of 120 soccer games was used to estimate the proportion (hat{p}) of home team wins.

The sample proportion is calculated by dividing the number of times the characteristic of interest is observed by the total number of observations in the sample. When calculating the bootstrap distribution, this sample proportion is used repeatedly as we simulate resampling from the population to create numerous new samples. This demonstrates the variability of the sample proportion and how it can serve as an estimate for the population proportion. Accurate use and interpretation of the sample proportion are essentials skills for students, as it can affect conclusions about the population based on their collected data.

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Most popular questions from this chapter

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

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?

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\) A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lcc} \hline \text { Case } & \text { Situation 1 } & \text { Situation 2 } \\ \hline 1 & 77 & 85 \\ 2 & 81 & 84 \\ 3 & 94 & 91 \\ 4 & 62 & 78 \\ 5 & 70 & 77 \\ 6 & 71 & 61 \\ 7 & 85 & 88 \\ 8 & 90 & 91 \\ \hline \end{array} $$

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

In Exercises 6.152 and \(6.153,\) find a \(95 \%\) confidence interval for the difference in proportions two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Difference in proportion who use text messaging, using \(\hat{p}_{t}=0.87\) with \(n=800\) for teens and \(\hat{p}_{a}=0.72\) with \(n=2252\) for adults.

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