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Find a \(95 \%\) confidence interval for the mean two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the t-distribution and the formula for standard error. Compare the results. Mean price of a used Mustang car online, in \$1000s, using data in MustangPrice with \(\bar{x}=15.98\), \(s=11.11,\) and \(n=25\)

Short Answer

Expert verified
Using the bootstrap method, the confidence interval is calculated using percentiles from the bootstrap distribution of the sample mean generated by resampling the data. On the other hand, using the t-distribution method, the 95% confidence interval for the given sample mean, standard deviation and size is approximately \(15.98 \pm 2.064 \times \frac{11.11}{\sqrt{25}}\). While the exact numerical answers would depend on the generated bootstrap distribution, the intervals from both methods should be fairly close most of the times, reflecting the variability in averages we would expect to see if we could draw many samples of Mustang car prices.

Step by step solution

01

Bootstrap Method

Bootstrap is a method for estimating parameters of the population distribution based on a random resampling of the sample data with replacement. To perform this method, we need to do the following steps: 1. Create a large number of pseudosamples by resampling from the original data with replacement. 2. Then for each pseudosample, compute the desired statistic (in this case, the mean). Since we don't have actual data points to generate the bootstrap distribution, we will assume the procedure is already done using some statistics software like StatKey or R. Usually, the 2.5 percentile and the 97.5 percentile of this distribution forms a 95% confidence interval.
02

t-Distribution Method

The t-distribution method involves applying inferential statistics theory. We need to use the following formula to calculate the 95% confidence interval: \(\bar{x} \pm t^{*} \frac{s}{\sqrt{n}}\). Where \(\bar{x}\) is the sample mean, \(s\) is the standard deviation, \(n\) is the sample size and \(t^{*}\) is the t-score from t-distribution table corresponding to desired confidence level and degrees of freedom (\(df = n - 1\)). Let's calculate it: The degrees of freedom is \(df = 25 - 1 = 24\). The t-score with 24 degrees of freedom for a two-tailed 95% confidence interval is approximately 2.064. Now we can plug these values into the formula to get the confidence interval: \(15.98 \pm 2.064 \times \frac{11.11}{\sqrt{25}}\).
03

Compare the results

Once we have calculated the 95% confidence interval using both methods, we compare those intervals. It's important to note that due to the random resampling in the bootstrap method, outcomes could be slightly different each time. The t-distribution method gives a more theoretical interval calculated based on the properties of t-distribution.

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Most popular questions from this chapter

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 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

In Exercises 6.188 to 6.191 , 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 \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

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 30 successes with \(n=100\) and Sample \(\mathrm{B}\) has a count of 50 successes with \(n=250\).

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 0.02 .

Use a t-distribution and the given matched pair 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 distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

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