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Use StatKey or other technology to generate a bootstrap distribution of sample 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 deviation as an estimate of the population standard deviation. Mean price of used Mustang cars online (in \$1000s) using the data in MustangPrice with \(n=25\), \(\bar{x}=15.98,\) and \(s=11.11\)

Short Answer

Expert verified
To determine the standard error, both the bootstrap method and Central Limit Theorem were used. After generating a bootstrap distribution of sample means, the standard error was calculated. According to Central Limit Theorem, the standard error of the distribution was computed to be 2.22. In practice, these two values should be similar if the sample is representative of the population.

Step by step solution

01

Calculate Bootstrap Standard Error

First, generate a bootstrap distribution of sample means. This would involve repeatedly resampling from the observed data, each time calculating the sample mean and storing it. Repeat this process many times, say 10,000, to generate a distribution. The standard deviation of this distribution is the bootstrap estimate of the standard error of the mean. In StatKey, this would be labeled as 'standard deviation of bootstrap distribution'.
02

Calculate Standard Error Using CLT

The Central Limit Theorem states that if you have a population with mean \(\mu\) and standard deviation \(\sigma\) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will approximate a normal distribution. The standard error can be calculated as \(\sigma / \sqrt{n}\), where \(n\) is the sample size and \(\sigma\) is the standard deviation of the population, estimated here by the sample standard deviation \(s\). In this case, the standard error would be \(11.11 / \sqrt{25} = 2.22\).
03

Compare Bootstrap And Theoretical Standard Errors

Now, compare the bootstrap standard error obtained in Step 1 with the theoretical one calculated in Step 2 based on Central Limit Theorem (CLT). This step involves simply comparing the numerical values. If the sample is representative of the population, the two values should be close.

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Most popular questions from this chapter

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.

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. 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. A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=501, s_{1}=115, n_{1}=400\) and \(\bar{x}_{2}=469, s_{2}=96, n_{2}=200 .\)

Football Air Pressure During the NFL's 2014 AFC championship game, officials measured the air pressure on game balls following a tip that one team's balls were under-inflated. In exercise 6.124 we found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under-inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of \(12.63,\) with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot's balls was significantly less than the average air pressure in the Indianapolis Colt's balls.

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 \(n=300\).

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.05 .

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