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In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a difference in proportions \(p_{1}-p_{2}\) if the samples have \(n_{1}=50\) with \(\hat{p}_{1}=0.68\) and \(n_{2}=80\) with \(\hat{p}_{2}=0.61,\) and the standard error is \(S E=0.085\)

Short Answer

Expert verified
The \(95\% \) confidence interval for the difference in proportions \( p_{1}-p_{2} \) is given by the range \(-0.0966 \leq p_{1}-p_{2} \leq 0.2366\)

Step by step solution

01

Identify the Z-score for the given confidence level

A \(95 \%\) confidence interval corresponds to an alpha level (\(\alpha\)) of 0.05. Since this is a two-tail test, divide \(\alpha\) by 2 to get 0.025. Looking this up in the Z-table, you'll find that the corresponding Z-score is approximately 1.96.
02

Apply the formula for the confidence interval

Plugging the values into the formula, the confidence interval for the difference in proportions can be found: \(( \hat{p}_{1} - \hat{p}_{2}) \pm Z * SE = (0.68 - 0.61) \pm 1.96 * 0.085 \)
03

Calculating the interval

Perform the above calculation to find the confidence interval: \(0.07 \pm 0.1666\) This results in the interval \( (-0.0966, 0.2366) \)

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Key Concepts

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

Bootstrap Distribution
The concept of bootstrap distribution is crucial when estimating confidence intervals in statistics. The idea is simple: we create many simulated samples from our original dataset to understand its variability.
Imagine you have a large dataset, and you take a small sample from it. The bootstrap method involves resampling this sample with replacement to create many 'bootstrap samples'. This technique helps to estimate the sampling distribution of a statistic, like the mean or proportion, without relying on theoretical distributions.
  • Resampling with Replacement: Each new sample is drawn from the original, and elements can be chosen more than once.
  • Simulating Real-World Variation: It provides a way to mimic the uncertainty of your estimate.
  • Usage: Particularly useful when the sample size is small or the population distribution is unknown.
This method gives us an approximate distribution of the sample statistics, such as means or proportions, that we can use to calculate confidence intervals.
Difference in Proportions
In statistics, sometimes we need to compare proportions between two independent groups. Let's say we have two groups: Group 1 with proportion \( \hat{p}_1 \) and Group 2 with proportion \( \hat{p}_2 \). The difference in proportions, \( \hat{p}_1 - \hat{p}_2 \), tells us how much larger or smaller one group’s proportion is compared to the other.
This difference is important, especially in hypothesis testing and confidence interval estimation. For example, you might want to compare customer satisfaction rates between two different stores.
  • Group Proportions: Calculated by dividing the number of successes in each group by the total number of observations in that group.
  • Interpretation: A positive difference indicates \( \hat{p}_1 \) is larger than \( \hat{p}_2 \), while a negative difference suggests the opposite.
  • Practical Use: Common in areas like medicine, market research, and social sciences to measure effect size or change.
This difference serves as the basis for calculating a confidence interval, allowing us to make inferences about the population means.
Standard Error
The standard error (SE) is a key concept in understanding the variability of an estimator. It measures how much the estimate of a statistic, such as a mean or proportion, would likely vary from one sample to another.
For the difference in proportions, the standard error is calculated using individual standard errors from each group. Having the SE allows us to understand the precision of the sample's estimate of the population parameter.
  • Formula: For proportions, SE is often found using the formula \( \sqrt{ \frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} } \).
  • Purpose: Helps in constructing confidence intervals and conducting hypothesis tests.
  • Influence: A smaller SE indicates a more precise estimate, which can provide more confidence in the results.
Understanding the standard error is crucial when interpreting the results of a study or research that involves estimating a population parameter.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations a data point is from the mean of a population. It is especially useful in hypothesis testing and constructing confidence intervals. A Z-score is used when the sampling distribution of the statistic is approximately normal.
In the context of confidence intervals, the Z-score is determined based on the chosen confidence level. A common choice is 95%, which corresponds to a Z-score of approximately 1.96 because it encompasses most values in a normal distribution.
  • Determination: For a two-tailed test with \( \alpha = 0.05 \), divide \( \alpha \) by 2 to get 0.025 for each tail.
  • Use in Bounds: Multiply the standard error by the Z-score to set the margin of error in a confidence interval.
  • Practical Significance: Helps to determine how extreme a particular observation is compared to the overall population distribution.
Calculating the Z-score provides a standardized way to understand the position of the data within the context of its distribution, crucial for making reliable inferences.

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Most popular questions from this chapter

Exercises 5.50 to 5.55 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) when the samples have \(n_{1}=n_{2}=50, \bar{x}_{1}=35.4, \bar{x}_{2}=33.1,\) \(s_{1}=1.28,\) and \(s_{2}=1.17 .\) The standard error of \(\bar{x}_{1}-\bar{x}_{2}\) from the randomization distribution is \(0.25 .\)

Exercises 5.21 to 5.28 ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves.] The lower \(10 \%\) for a standard normal distribution converted to a \(N(500,80)\) distribution.

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{11}\) in Scotland both before and after comprehensive smoke-free legislation was passed in March \(2006 .\) Monthly records were kept of the annualized percent change in asthma admissions, both before and after the legislation was passed. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of 5.2\% per year. The standard error for this estimate is \(0.7 \%\) per year. After the legislation, admissions were decreasing at a mean rate of \(18.2 \%\) per year, with a standard error for this mean of \(1.79 \% . \mathrm{In}\) both cases, the sample size is large enough to use a normal distribution. (a) Find and interpret a \(95 \%\) confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation. (b) Find a \(95 \%\) confidence interval for the same quantity after the legislation. (c) Is this an experiment or an observational study? (d) The evidence is quite compelling. Can we conclude cause and effect?

Empirical Rule for Normal Distributions Pick any positive values for the mean and the standard deviation of a normal distribution. Use your selection of a normal distribution to answer the questions below. The results of parts (a) to (c) form what is often called the Empirical Rule for the standard deviation in a normal distribution. (a) Verify that about \(95 \%\) of the values fall within two standard deviations of the mean. (b) What proportion of values fall within one standard deviation of the mean? (c) What proportion of values fall within three standard deviations of the mean? (d) Will the answers to (b) and (c) be the same for any normal distribution? Explain why or why not.

(a) The area below 80 on a \(N(75,10)\) distribution (b) The area above 25 on a \(N(20,6)\) distribution (c) The area between 11 and 14 on a \(N(12.2,1.6)\) distribution

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