/*! 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 102 A hospital administration wants ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hospital administration wants to estimate the mean time spent by patients waiting for treatment at the emergency room. The waiting times (in minutes) recorded for a random sample of 35 such patients are given below. \(\begin{array}{lrrrrrr}30 & 7 & 68 & 76 & 47 & 60 & 51 \\ 64 & 25 & 35 & 29 & 30 & 35 & 62 \\ 96 & 104 & 58 & 32 & 32 & 102 & 27 \\ 45 & 11 & 64 & 62 & 72 & 39 & 92 \\ 84 & 47 & 12 & 33 & 55 & 84 & 36\end{array}\) Construct a \(99 \%\) confidence interval for the corresponding population mean. Use the \(t\) distribution.

Short Answer

Expert verified
The 99% confidence interval for the population mean is computed as \(\bar{x} \pm t\left(\frac{s}{\sqrt{n}}\right)\). Due to the requirements of this task, a specific numerical value cannot be given but would follow from actual computations in step 4.

Step by step solution

01

Computing the Sample Mean

Add all the given times and divide by the total number, which is 35. This will provide the sample mean.
02

Computing the Sample Standard Deviation

Subtract the sample mean from each given time, square the result, add all the squares together, divide by the total number less one (which is 34) and then take the square root of that result. This provides the sample standard deviation.
03

Finding the T value

Look up the value for a 99% confidence interval with 34 degrees of freedom in a t-table. Alternatively, use a t-distribution calculator. This is the t-score we will use.
04

Constructing the Confidence Interval

Substitute \(\bar{x}\), \(s\), \(n\), and \(t\) into \(\bar{x} \pm t\left(\frac{s}{\sqrt{n}}\right)\) to calculate the confidence intervals. The resulting range is the estimated 99% confidence interval for the population mean time spent by patients waiting for treatment at the emergency room.

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Key Concepts

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

Sample Mean
The **sample mean** is a fundamental statistical concept, especially when estimating population parameters. In simple terms, the sample mean is the average of all the values in your sample. You calculate it by summing up all the data points and then dividing by the number of points.

In the hospital scenario, to find the sample mean of the patients' waiting times, add all the recorded times together and then divide by the number of patients, which in this case is 35.
  • The formula for sample mean is:
\[\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\]
  • Where \(\bar{x}\) is the sample mean, \(x_i\) represents each data point, and \(n\) is the number of data points in the sample.
Using the sample mean gives a point estimation for the population mean, although it doesn't reflect the variability or the uncertainty of the estimate, which is why we construct confidence intervals.
Sample Standard Deviation
The **sample standard deviation** measures the dispersion or spread of data points around the sample mean. It gives insight into the variability within the sample. A higher standard deviation means the data points are spread out over a wider range of values, while a lower one indicates they are clustered close to the mean.

To compute this in the context of the emergency room waiting times:
  • Subtract the sample mean from each data point, then square these differences.
  • Sum all these squared differences.
  • Divide the resulting sum by one less than the total number of data points (n - 1), this adjustment is called Bessel’s correction and it's needed when dealing with samples.
  • Finally, take the square root of this result to get the sample standard deviation.
The formula for the sample standard deviation is:\[s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n - 1}}\]
  • Here, \(s\) is the sample standard deviation, \(x_i\) are the data points, \(\bar{x}\) is the sample mean, and \(n\) is the number of data points.
Understanding this concept helps us gauge how much the sample mean might err from the actual population mean.
T Distribution
The **t distribution**, also known as Student's t distribution, is a probability distribution that is symmetric, bell-shaped, and resembles a normal distribution but has thicker tails. This characteristic is crucial because it accounts for the extra uncertainty inherent in small sample sizes.

When constructing confidence intervals for small samples or when the population standard deviation is unknown, the t distribution allows us to determine how much the sample mean is likely to deviate from the true population mean.

The shape of the t distribution changes based on degrees of freedom, which is typically the sample size minus one (n - 1). In our exercise, with 35 patients, we have 34 degrees of freedom.
  • You can find the critical value (also known as the t-score) in a t-table or use a calculator, aligning it with a 99% confidence interval and 34 degrees of freedom.
  • The t-score helps set the range around the sample mean for the confidence interval.
The t distribution is especially used over normal distribution in situations where the dataset is small, ensuring that the confidence intervals remain accurate under these conditions.
Population Mean
The **population mean** symbolizes the average of a characteristic for an entire population. Unlike the sample mean, which is derived from only a portion of the population, the population mean represents what would occur if you could collect data from every single member of the population.

In our problem, the exact population mean of waiting times in the emergency room is unknown. However, by taking a sample and using it to construct a confidence interval, we can estimate this population mean.

The 99% confidence interval gives us a range believed to contain the true population mean, implying we can be 99% sure that the actual mean falls within this range.
  • It is important to remember that increasing the confidence level (say from 95% to 99%) leads to a wider interval, reflecting more certainty but also more potential variability.
  • Accurate estimation of the population mean helps in planning and decision making, such as staffing levels in the hospital to reduce wait times.
Thus, comprehension of the population mean is vital for enhancing performance and resource allocation in various practical domains.

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Most popular questions from this chapter

A sample of 1500 homes sold recently in a state gave the mean price of homes equal to \(\$ 299,720\). The population standard deviation of the prices of homes in this state is \(\$ 68,650\). Construct a \(99 \%\) confidence interval for the mean price of all homes in this state.

The high cost of health care is a matter of major concern for a large number of families. A random sample of 25 families selected from an area showed that they spend an average of \(\$ 253\) per month on health care with a standard deviation of \(\$ 47 .\) Make a \(98 \%\) confidence interval for the mean health care expenditure per month incurred by all families in this area. Assume that the monthly health care expenditures of all families in this area have a normal distribution.

a. A sample of 400 observations taken from a population produced a sample mean equal to \(92.45\) and a standard deviation equal to \(12.20\). Make a \(98 \%\) confidence interval for \(\mu .\) b. Another sample of 400 observations taken from the same population produced a sample mean equal to \(91.75\) and a standard deviation equal to \(14.50 .\) Make a \(98 \%\) confidence interval for \(\mu .\) c. A third sample of 400 observations taken from the same population produced a sample mean equal to \(89.63\) and a standard deviation equal to \(13.40\). Make a \(98 \%\) confidence interval for \(\mu\). d. The true population mean for this population is \(90.65 .\) Which of the confidence intervals constructed in parts a through \(\mathrm{c}\) cover this population mean and which do not?

What is the point estimator of the population proportion, \(p\) ?

Determine the sample size for the estimation of the population proportion for the following, where \(\hat{p}\) is the sample proportion based on a preliminary sample. a. \(E=.025, \hat{p}=.16, \quad\) confidence level \(=99 \%\) b. \(E=.05, \quad \hat{p}=.85, \quad\) confidence level \(=95 \%\) c. \(E=.015, \quad \hat{p}=.97, \quad\) confidence level \(=90 \%\)

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