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Ch. 7 - Estimating Parameters and Determining Sample Sizes
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 7, Problem 7.3.21

Large Data Sets from Appendix B. In Exercises 21 and 22, use the data set in Appendix B. Assume that each sample is a simple random sample obtained from a population with a normal distribution.


Comparing Waiting Lines Refer to Data Set 30 “Queues” in Appendix B. Construct separate 95% confidence interval estimates of using the two-line wait times and the single-line wait times. Do the results support the expectation that the single line has less variation? Do the wait times from both line configurations satisfy the requirements for confidence interval estimates of sigma

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Step 1: Identify the data set and variables. From the problem, we are working with Data Set 30 'Queues' in Appendix B. The two variables of interest are the wait times for the two-line configuration and the single-line configuration. These are the two groups for which we will construct separate 95% confidence intervals for the population standard deviation (σ).
Step 2: Verify the assumptions. The problem states that the samples are simple random samples and the population has a normal distribution. Confirm that the wait times for both configurations satisfy these assumptions by checking the data for normality (e.g., using a histogram, Q-Q plot, or a normality test like the Shapiro-Wilk test).
Step 3: Use the chi-square distribution to construct confidence intervals for the population standard deviation. The formula for the confidence interval of σ is: \( \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}} \right) \), where \( n \) is the sample size, \( s \) is the sample standard deviation, and \( \chi^2 \) are the critical values of the chi-square distribution with \( n-1 \) degrees of freedom.
Step 4: Calculate the sample standard deviation (s) for each group (two-line and single-line wait times) and determine the sample size (n). Use these values to compute the confidence intervals for each group using the formula from Step 3. Look up the critical chi-square values for a 95% confidence level and \( n-1 \) degrees of freedom.
Step 5: Compare the confidence intervals for the two groups. If the confidence interval for the single-line configuration is narrower (indicating less variation), this supports the expectation that the single line has less variation. Additionally, confirm that the wait times satisfy the requirements for confidence interval estimates of σ by ensuring the data is approximately normal and the sample sizes are sufficient.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Confidence Interval

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified level of confidence, typically 95%. It provides an estimate of uncertainty around a sample mean or proportion, allowing researchers to infer about the population. The width of the interval reflects the variability in the data and the sample size, with larger samples generally yielding narrower intervals.
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Variation and Standard Deviation

Variation refers to how much data points in a set differ from each other and from the mean. Standard deviation is a key measure of this variation, quantifying the average distance of each data point from the mean. In the context of waiting times, lower variation indicates more consistent wait times, which is crucial for comparing different queue configurations, such as single-line versus two-line systems.
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Normal Distribution

A normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. Many statistical methods, including confidence intervals, assume that the underlying data follows a normal distribution. This assumption is important when analyzing wait times, as it affects the validity of the confidence intervals constructed from the sample data.
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Related Practice
Textbook Question

In Exercises 5–8, (a) identify the critical value ta/2 used for finding the margin of error, (b) find the margin of error, (c) find the confidence interval estimate of u, and (d) write a brief statement that interprets the confidence interval.


Pepsi Weights Here are summary statistics for the weights of Pepsi in randomly selected cans: n=36, x=0.82410 lb, s=0.00570 lb (based on Data Set 37 “Cola Weights and Volumes” in Appendix B). Use a confidence level of 99%.

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Textbook Question

Constructing and Interpreting Confidence Intervals. In Exercises 13–16, use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion p; (b) identify the value of the margin of error E; (c) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval.


Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Construct a 95% confidence interval for the proportion of medical malpractice lawsuits that are dropped or dismissed.

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Textbook Question

Formats of Confidence Intervals. In Exercises 9–12, express the confidence interval using the indicated format. (The confidence intervals are based on the proportions of red, orange, yellow, and blue M&Ms in Data Set 38 “Candies” in Appendix B.)


Green M&Ms Express 0.116 < p < 0.192 in the form of p +-E.

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Textbook Question

Body Temperature Data Set 5 “Body Temperatures” in Appendix B includes a sample of 106 body temperatures having a mean of and a standard deviation of 0.62F (for day 2 at 12 AM). Construct a 95% confidence interval estimate of the standard deviation of the body temperatures for the entire population.

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Textbook Question

Mercury in Sushi An FDA guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in New York City. The study was sponsored by the New York Times, and the stores (in order) are D’Agostino, Eli’s Manhattan, Fairway, Food Emporium, Gourmet Garage, Grace’s Marketplace, and Whole Foods. Construct a 98% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi?


0.56 0.75 0.10 0.95 1.25 0.54 0.88

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Textbook Question

Sample Size for Proportion Find the sample size required to estimate the percentage of statistics students who take their statistics course online. Assume that we want 95% confidence that the proportion from the sample is within two percentage points of the true population percentage.

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