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Chapter 6: Problem 32

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion 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. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

Short Answer

Expert verified
The confidence intervals using bootstrap distribution and normal distribution should provide similar results due to the central limit theorem, which states that, with a large enough sample size, the sampling distribution of the mean will be normally distributed. Any minor discrepancy could occur due to approximation when applying the normal distribution or randomness in bootstrap samples, but in general, they should more or less align.

Step by step solution

01

Find Confidence Interval using StatKey or other Technology and Bootstrap Distribution

Use StatKey or other appropriate software to create bootstrap samples and draw a corresponding bootstrap distribution. Then, extract percentiles from the bootstrap distribution to obtain the confidence interval. For a 95% confidence interval, extract the 2.5th percentile and the 97.5th percentile. These represent the lower and upper limits of the confidence interval, respectively.
02

Find Confidence Interval Using Normal Distribution and Standard Error

To calculate the confidence interval using the normal distribution, first find the standard error. The formula for standard error, \(SE\), when dealing with proportions is \(\sqrt{\hat{p}(1-\hat{p})/n}\). Substituting the given, we have \(SE = \sqrt{0.583(1-0.583)/120}\). Next, we find the Z-score that corresponds to a 95% confidence interval, which is approximately 1.96. Then, for the lower limit we calculate \(\hat{p} - Z*SE\) and for the upper limit, we calculate \(\hat{p} + Z*SE\).
03

Compare the Results

Upon receiving both confidence intervals, compare them to verify the similarity. Differences could occur due to the approximation when using the normal distribution and the randomness when using bootstrap samples. They both, however, should give pretty close results.

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Key Concepts

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

Bootstrap Distribution
The bootstrap distribution is a powerful tool used to estimate the sampling distribution of a statistic by resampling with replacement from the original data set. This method is particularly useful because it does not assume a specific distribution for the sample, making it versatile.
  • Bootstrap involves generating many samples, often thousands, from the original data set.
  • Each sample is obtained by randomly selecting data points, with replacement, from the data set. This means some points might be picked multiple times, while some not at all.
  • The statistic of interest, such as a mean or proportion, is calculated from these resamples.
  • The collection of these statistics forms the bootstrap distribution.
To find a confidence interval using bootstrap distributions, we look at the percentiles of the distribution. For example, in a 95% confidence interval, the 2.5th percentile serves as the lower limit and the 97.5th percentile as the upper limit. This gives us a range where we expect the true parameter to lie with high confidence.
Standard Error
The standard error (SE) is a measure of the variability in a sample statistic, and it provides an estimate of how far the sample statistic is likely to be from the population parameter. When calculating a confidence interval, SE is crucial as it helps determine the width of the interval.
  • For proportions, the formula is \[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]where \(\hat{p}\) is the sample proportion and \(n\) is the sample size.
  • This formula indicates that as the sample size increases, the standard error decreases, meaning our estimate becomes more precise.
  • Standard error allows for the calculation of a margin of error, which is added and subtracted from the sample proportion to create the confidence interval.
In contexts like the exercise, understanding and calculating SE is essential to apply the normal distribution method for constructing confidence intervals accurately.
Normal Distribution
The normal distribution is a fundamental concept in statistics and is often used for calculations involving probabilities and sampling distributions due to its symmetric, bell-shaped curve.
  • It's characterized by two parameters: the mean (where the peak is located) and the standard deviation (which dictates the spread).
  • Many statistical techniques, including confidence interval estimation, presume that the data or certain sample statistics will follow a normal distribution due to the Central Limit Theorem.
  • When calculating confidence intervals using a normal distribution, we apply the Z-score, which is a measure of how many standard deviations an element is from the mean.
For a 95% confidence interval, the Z-score of approximately 1.96 is utilized, representing the range within which 95% of sample statistics fall under the normal curve. This is pivotal when applying the normal approximation method to determine the probability or confidence interval of proportions as shown in the given exercise.

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Most popular questions from this chapter

Exercises 6.192 and 6.193 examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DEHP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=83.6\) with \(s_{F}=194.7\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=59.1\) with \(s_{N}=152.1\)

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 \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Who Exercises More: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent exercising in a typical week. Computer output of descriptive statistics for the number of hours spent exercising, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: Exercise } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\\ \text { Exercise } & \mathrm{F} & 168 & 8.110 & 5.199 \\ & \mathrm{M} & 193 & 9.876 & 6.069\end{array}\) \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.000 & 4.000 & 7.000 & 12.000 & 27.000 \\\ 0.000 & 5.000 & 10.000 & 14.000 & 40.000\end{array}\) (a) How many females are in the dataset? How many males? (b) In the sample, which group exercises more, on average? By how much? (c) Use the summary statistics to compute a \(95 \%\) confidence interval for the difference in mean number of hours spent exercising. Be sure to define any parameters you are estimating. (d) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: Two-sample \(\mathrm{T}\) for Exercise Gender N Mean StDev SE Mean \(\begin{array}{lllll}\mathrm{F} & 168 & 8.11 & 5.20 & 0.40 \\ \mathrm{M} & 193 & 9.88 & 6.07 & 0.44\end{array}\) Difference \(=\operatorname{mu}(F)-\operatorname{mu}(M)\) Estimate for difference: -1.766 \(95 \%\) Cl for difference: (-2.932,-0.599)

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 \(n=25\)

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 \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=10.1, s_{1}=2.3, n_{1}=50\) and \(\bar{x}_{2}=12.4, s_{2}=5.7, n_{2}=50 .\)
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