/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 152 You hurry to the local emergency... [FREE SOLUTION] | 91Ó°ÊÓ

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You hurry to the local emergency department in hopes of immediate, urgent care, only to find yourself waiting for what seems like hours. The manager of the large emergency department believes that his new procedures have substantially reduced the wait time for the average urgent care patient. He initiates a study to evaluate the wait time. The records of 18 randomly selected patients seen since the new procedures were put in place are checked, and the time between entering the emergency department and being seen by urgent care personnel was observed. The mean wait time was 17.82 minutes with a standard deviation of 5.68 minutes. Estimate the mean wait time using a \(99 \%\) confidence interval. Assume that wait times are normally distributed.

Short Answer

Expert verified
The estimated mean wait time, using a 99% confidence interval, is between 14.33 minutes and 21.31 minutes.

Step by step solution

01

Identify given statistics

In this exercise, there is a sample size of \( n = 18 \) patients, a sample mean \( \overline{X} \) of 17.82 minutes, and a standard deviation \( s \) of 5.68 minutes. A 99% confidence interval is desired, which corresponds to an alpha level of \( \alpha = 0.01 \).
02

Determine the z-score associated with the desired level of confidence

Since the level of confidence is 99%, the alpha level is divided by 2, resulting in 0.005 (since the z-score represents the extreme ends of both tails of the normal distribution). So the z-value associated with \( \alpha/2 = 0.005 \) is approximately 2.58.
03

Apply the formula for the confidence interval

The formula for a confidence interval for a population mean is \( \overline{X} \pm Z \times \frac{s}{\sqrt{n}} \). Plugging in the given figures: Mean \( \overline{X}= 17.82 \), Z-value= 2.58, standard deviation \( s = 5.68 \), and sample size \( n = 18 \), the calculation becomes \( 17.82 \pm 2.58 \times \frac{5.68}{\sqrt{18}} \).
04

Compute the interval's upper and lower limits

Solving the calculation from Step 3 provides the upper and lower bounds of the interval. \( 17.82 + 2.58 \times \frac{5.68}{\sqrt{18}} = 21.31 \) (upper limit) \( 17.82 - 2.58 \times \frac{5.68}{\sqrt{18}} = 14.33 \) (lower limit)

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Key Concepts

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

Normal Distribution
In statistics, the **normal distribution** is a key concept that describes how data tends to cluster around a central value in many natural phenomena. Imagine a bell curve: symmetrical and peaking at the mean. This is the hallmark of a normal distribution. It means most data points are close to the average, with fewer points farther away.
Such distribution appears in countless scenarios. For example, human heights, test scores, and, as in our exercise, wait times in an emergency department. This characteristic makes the normal distribution a vital tool in statistics.
When we say wait times are "normally distributed," we're assuming they follow this bell curve shape. This assumption is crucial because it allows statisticians to apply various statistical tools, including confidence intervals, to make inferences about populations based on sample data.
Sample Mean
The **sample mean** is the average of the data collected from a sample. It's calculated by summing up all the observed values and dividing them by the number of observations. In our scenario, the sample mean is 17.82 minutes.
We use sample means to estimate the population mean, which represents the average wait time for all patients. After all, studying the entire population is often impractical. That's why we rely on samples to give us a glimpse of the larger picture.
It's important to note that the sample mean is an approximation of the true population mean. This approximation is worthwhile when the sample is representative of the population, meaning its characteristics mirror those of the larger group.
Standard Deviation
The **standard deviation** is a measure of how spread out the values in a dataset are. It tells us how much individual data points differ from the mean. In our exercise, the standard deviation is 5.68 minutes.
A smaller standard deviation means the data points are closer to the mean. Conversely, a larger standard deviation suggests more variability. This variability is significant as it helps us assess the reliability of the sample mean.
By understanding the standard deviation, we can gauge how concentrated the wait times are around the average. This is crucial when estimating confidence intervals, which depend on the spread of the data to make accurate predictions.
Z-Score
The **z-score** is a statistical tool that measures the distance of a data point from the population mean in terms of standard deviations. In this exercise, the z-score helps establish a confidence interval for the population mean wait time.
For a 99% confidence level, the z-score is approximately 2.58. This value can be found using z-tables or statistical software, marking how many standard deviations away from the mean are required to include 99% of the data.
The z-score is essential in calculating the margin of error in a confidence interval. It allows us to account for the distribution's spread in a mathematically sound manner. By multiplying it with the standard deviation and dividing by the square root of the sample size, we determine the range in which we expect the true mean to fall.

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Most popular questions from this chapter

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

For a chi-square distribution having 35 degrees of freedom, find the area under the curve between \(\chi^{2}(35,0.96)\) and \(\chi^{2}(35,0.15).\)

Three nationwide poll results are described below. USA Today Snapshot/Rent.com, August 18,2009 \(N=1000\) adults 18 and over; \(\mathrm{MoE} \pm 3 .\) (MoE is margin of error. "What renters look for the most when seeking an apartment:" Washer/dryer\(-39\%,\) Air Conditioning \(-30 \%,\) Fitness Center- \(10 \%,\) Pool \(-10 \%\) USA Today/Harris Interactive Poll, February \(10-15,2009 ; N=1010\) adults; MoE ±3. "Americans who say people on Wall Street are "as honest and moral as other people." Disagree \(-70 \%\) Agree \(-26 \%,\) Not sure/refuse to answer \(-4 \%\) American Association of Retired Persons Bulletin/AARP survey, July 22-August 2, 2009; \(N=1006\) adults age 50 and older; \(\mathrm{MoE} \pm 3\). The American Association of Retired Persons Bulletin Survey reported that \(16 \%\) of adults, 50 and older, said they are likely to return to school. Each of the polls is based on approximately 1005 randomly selected adults. a. Calculate the \(95 \%\) confidence maximum error of estimate for the true binomial proportion based on binomial experiments with the same sample size and observed proportion as listed first in each article. b. Explain what caused the values of the maximum errors to vary. c. The margin of error being reported is typically the value of the maximum error rounded to the next larger whole percentage. Do your results in part a verify this? d. Explain why the round-up practice is considered "conservative." e. What value of \(p\) should be used to calculate the standard error if the most conservative margin of error is desired?

a. What value of chi-square for 5 degrees of freedom subdivides the area under the distribution curve such that \(5 \%\) is to the right and \(95 \%\) is to the left? b. What is the value of the 95 th percentile for the chi-square distribution with 5 degrees of freedom? c. What is the value of the 90th percentile for the chi-square distribution with 5 degrees of freedom?

a. The central \(90 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? b. The central \(95 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? c. The central \(99 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values?

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