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Use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting distance (in miles) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=18.16,\) and \(s_{1}=13.80\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=14.16,\) and \(s_{2}=10.75\) for St. Louis.

Short Answer

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The first step in the solution involves calculating the difference in means which results in 4 miles. After that, by inputting given numbers into the formula of standard error according to the Central Limit Theorem, one can obtain a standard error. This then needs to be compared with the bootstrap standard error, which could have been calculated if the raw data were provided.

Step by step solution

01

Bootstrapping

Bootstrapping involves drawing repeated samples from the observed data and recalculating the metric of interest. However, in this problem scenario, the raw data is not provided; hence, direct bootstrapping cannot be performed. Assuming the need to find the theoretical bootstrap standard error, it can be found by calculating the standard deviation of the difference in means.
02

Calculate Difference in Mean

The difference in means for two samples is given as: \( \Delta \mu = \bar{x}_{1} - \bar{x}_{2} \). Substituting the values given, we get \(\Delta \mu = 18.16 - 14.16 = 4\). So, the difference in means is 4 miles.
03

Calculate Standard Error - CLT

The standard error of the difference in means, according to the CLT, can be calculated using the formula: \(SE_{\Delta \mu} = \sqrt{ \frac{s^2_{1}}{n_{1}} + \frac{s^2_{2}}{n_{2}}}\). Substituting given values into the equation: \(SE_{\Delta \mu} = \sqrt{ \frac{13.80^2}{500} + \frac{10.75^2}{500}}\). After performing the calculation one can obtain the standard error according to the Central Limit Theorem.
04

Comparison

Finally, compare the bootstrap standard error with the SE calculated from the Central Limit Theorem.

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Most popular questions from this chapter

Dark Chocolate for Good Health A study \(^{47}\) examines chocolate's effects on blood vessel function in healthy people. In the randomized, doubleblind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of \(2.32,\) while the control group had a mean change of -0.96 with a standard deviation of 1.58 . (a) Explain what "randomized, double-blind, placebo-controlled study" means. (b) Find and interpret a \(95 \%\) confidence interval for the difference in means between the two groups. Be sure to clearly define the parameters you are estimating. You may assume that neither sample shows significant departures from normality. (c) Is it plausible that there is "no difference" between the two kinds of chocolate? Justify your answer using the confidence interval found in \(\operatorname{part}(\mathrm{b})\)

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Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Bananas after 15 Days After 15 days, 345 of the 500 fruit flies eating organic bananas are still alive, while 320 of the 500 eating conventional bananas are still alive.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\).

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the sample results \(\bar{x}_{1}=56, s_{1}=8.2\) with \(n_{1}=30\) and \(\bar{x}_{2}=51, s_{2}=6.9\) with \(n_{2}=40\).

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