/*! 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 16 What percentage of hospitals pro... [FREE SOLUTION] | 91Ó°ÊÓ

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What percentage of hospitals provide at least some charity care? The following problem is based on information taken from State Health Care Data: Utilization, Spending, and Characteristics (American Medical Association). Based on a random sample of hospital reports from eastern states, the following information was obtained (units in percentage of hospitals providing at least some charity care): \(\begin{array}{llllllllll}57.1 & 56.2 & 53.0 & 66.1 & 59.0 & 64.7 & 70.1 & 64.7 & 53.5 & 78.2\end{array}\) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 62.3 \%\) and \(s \approx 8.0 \%\). Find a \(90 \%\) confidence interval for the population average \(\mu\) of the percentage of hospitals providing at least some charity care.

Short Answer

Expert verified
The 90% confidence interval for the mean percentage is approximately \([57.66\%, 66.94\%]\).

Step by step solution

01

Understand the Problem Requirements

We are given a random sample of percentages of hospitals providing charity care and need to find a 90% confidence interval for the population mean, \( \mu \). A confidence interval gives a range within which the true population mean is likely to lie.
02

Review Provided Data

The given data \( \{57.1, 56.2, 53.0, 66.1, 59.0, 64.7, 70.1, 64.7, 53.5, 78.2\} \) is a sample of hospital percentages. We have the sample mean \( \bar{x} \approx 62.3\% \) and sample standard deviation \( s \approx 8.0\% \) provided for these values.
03

Calculate Sample Size

First, find the sample size \( n \). Count the number of observations in the sample: \( n = 10 \).
04

Find the Critical Value for 90% Confidence Level

Since the sample size is small (\( n = 10 \)), use the t-distribution. The degrees of freedom \( df = n - 1 = 9 \). For a 90% confidence level and \( df = 9 \), use a t-table to find \( t_{\alpha/2} \approx 1.833 \).
05

Calculate the Standard Error of the Mean

The standard error (SE) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{8.0}{\sqrt{10}} \approx 2.53 \].
06

Determine the Confidence Interval

Calculate the margin of error (MOE): \[ MOE = t_{\alpha/2} \times SE = 1.833 \times 2.53 \approx 4.64 \]. The confidence interval is then \( \bar{x} \pm MOE = 62.3 \pm 4.64 \).
07

Present the Confidence Interval

Calculate the lower and upper bounds of the confidence interval: Lower bound \( = 62.3 - 4.64 \approx 57.66\% \) and upper bound \( = 62.3 + 4.64 \approx 66.94\% \). Therefore, the 90% confidence interval is approximately \([57.66\%, 66.94\%]\).

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Key Concepts

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

Confidence Interval
A confidence interval is a statistical tool that provides a range of values that is likely to contain a population parameter with a certain level of confidence. In the context of our exercise, we want to find out how confident we can be about the true average percentage of hospitals providing charity care based on our sample data.
When calculating a confidence interval, you choose a confidence level, such as 90%, 95%, or 99%, representing how sure you are that the true parameter lies within this interval. For example, a 90% confidence interval means that if we were to take many samples and calculate the confidence interval for each, about 90% of these intervals would contain the true population mean.
The steps involved in finding a confidence interval include calculating the sample mean, determining the standard error, and considering the variability of the data through a critical value from the t-distribution, then combining these elements to determine the margin of error. Finally, adding and subtracting this margin of error from the sample mean gives us the confidence interval.
Sample Mean
The sample mean is an average of the data points in a sample and serves as an estimate of the population mean. It is denoted as \( \bar{x} \). In this exercise, the sample mean helps us summarize the central tendency of the hospital percentages in our sample.
To calculate the sample mean, you add up all the sample values and divide by the number of observations in the sample. This gives a single value that represents the average of your data points. In our case, the sample mean is approximately 62.3%, which tells us that on average, around 62.3% of hospitals in the sample provide some charity care.
Using the sample mean is crucial because it acts as the starting point for further statistical analysis, including constructing the confidence interval. It allows for a way to infer about the entire population from which the sample is drawn.
T-Distribution
The t-distribution is a key concept when working with small sample sizes, especially when the population standard deviation is unknown. Unlike the normal distribution, the t-distribution has thicker tails, meaning it accounts for more variability in smaller samples.
In the given problem, because our sample size is 10, we use the t-distribution. Its shape is determined by the degrees of freedom, calculated as the number of sample observations minus one (\( n - 1 \)). Here, we have 9 degrees of freedom. Using these, we determine a critical value (\( t_{\alpha/2} \)) from a t-table, which helps us find the margin of error for our confidence interval.
The t-distribution becomes very similar to the normal distribution as the sample size increases. For small samples, like ours, it provides a more accurate critical value due to the extra variability expected in smaller datasets. This adjustment helps ensure our confidence interval is correctly estimated.
Standard Error
The standard error (SE) measures the variability or dispersion of the sample mean from the true population mean. It is an essential element in estimating a confidence interval, letting us understand how much our sample mean would vary if we were to take multiple samples from the same population.
To calculate the standard error, we divide the sample standard deviation \( s \) by the square root of the sample size \( n \). In our problem, the formula is: \[ SE = \frac{s}{\sqrt{n}} = \frac{8.0}{\sqrt{10}} \approx 2.53 \]A smaller standard error indicates that the sample mean is a more accurate estimate of the population mean. Conversely, a larger standard error suggests more spread and potentially more variability, making it harder to pin down the population mean precisely.
In constructing the confidence interval, the standard error combines with the critical value from the t-distribution to calculate the margin of error, determining the range of the interval. This process makes the standard error a critical component in the reliability and accuracy of our confidence estimates.

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Most popular questions from this chapter

You want to conduct a survey to determine the proportion of people who favor a proposed tax policy. How does increasing the sample size affect the size of the margin of error?

Thirty small communities in Connecticut (population near 10,000 each) gave an average of \(\bar{x}=138.5\) reported cases of larceny per year. Assume that \(\sigma\) is known to be \(42.6\) cases per year (Reference: Crime in the United States, Federal Bureau of Investigation). (a) Find a \(90 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (b) Find a \(95 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (c) Find a \(99 \%\) confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?

Sam computed a \(95 \%\) confidence interval for \(\mu\) from a specific random sample. His confidence interval was \(10.1<\mu<12.2 .\) He claims that the probability that \(\mu\) is in this interval is \(0.95\). What is wrong with his claim?

Consider a \(90 \%\) confidence interval for \(\mu .\) Assume \(\sigma\) is not known. For which sample size, \(n=10\) or \(n=20\), is the critical value \(t_{c}\) larger?

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