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Chapter 8: Problem 58

The administrators for a hospital wished to estimate the average number of days required for inpatient treatment of patients between the ages of 25 and \(34 .\) A random sample of 500 hospital patients between these ages produced a mean and standard deviation equal to 5.4 and 3.1 days, respectively. Construct a 95\% confidence interval for the mean length of stay for the population of patients from which the sample was drawn.

Short Answer

Expert verified
The 95% confidence interval is approximately [5.129, 5.671].

Step by step solution

01

Identify the Known Values

We need to construct a 95% confidence interval for the mean. From the problem, we have:- Sample size, \(n = 500\)- Sample mean, \( \bar{x} = 5.4 \) days- Standard deviation, \( s = 3.1 \) days.
02

Determine the Critical Value

For a 95% confidence interval, we use the standard normal distribution because the sample size is large (\(n = 500\)). The critical value (\(z\)) for a 95% confidence level is 1.96.
03

Calculate the Standard Error

The standard error (SE) of the mean is found using the formula:\[SE = \frac{s}{\sqrt{n}}\]Substitute the known values:\[SE = \frac{3.1}{\sqrt{500}} \approx 0.1384\]
04

Compute the Margin of Error

The margin of error (ME) is calculated using the formula:\[ME = z \times SE\]Substitute the critical value and standard error:\[ME = 1.96 \times 0.1384 \approx 0.2713\]
05

Construct the Confidence Interval

The confidence interval (CI) is found using the formula:\[\bar{x} \pm ME\]Substitute the mean and margin of error:\[5.4 \pm 0.2713\]Thus, the 95% confidence interval is approximately \([5.129, 5.671]\).

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Key Concepts

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

Sample Size
When you perform statistical analysis, the sample size, denoted as \( n \), is incredibly important for calculating a confidence interval. It refers to the number of observations or data points you collect from a population subset. In our exercise, the sample size is 500 patients.

Here's why sample size matters:
  • A larger sample size generally provides a more accurate estimate of the population parameter.
  • It reduces sampling error, which is the discrepancy between the sample statistic and the actual population parameter.
  • Increased sample sizes often lead to narrower confidence intervals, indicating more precise estimates.
Therefore, having a large sample size, like 500, helps in obtaining a reliable estimate of the average hospital stay duration for the population between ages 25 and 34.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of our exercise, the standard deviation \( s \) is 3.1 days.

Here's how standard deviation functions:
  • A low standard deviation means that the data points tend to be close to the mean, indicating less spread.
  • A high standard deviation means that the data points are spread out over a larger range of values, suggesting greater variability.
  • Standard deviation is crucial in calculating other statistical metrics, such as the standard error and confidence intervals.
Understanding how much variability exists in our sample (3.1 days, in this case) helps assess the confidence we have in the mean as a representative of the population.
Standard Error
The standard error (SE) of the mean provides insight into how much the sample mean \( \bar{x} \) is expected to differ from the actual population mean due to random sampling variability. It is calculated by the formula:\[SE = \frac{s}{\sqrt{n}}\]In the context of the original exercise, the standard error is approximately 0.1384 days.

This metric serves several purposes:
  • It quantifies the uncertainty associated with the sample mean as an estimate of the population mean.
  • The smaller the standard error, the more precise is the estimate of the population mean.
The concept of standard error helps when constructing confidence intervals, as it plays a critical role in defining the width of the interval.
Margin of Error
The margin of error (ME) provides a range that expresses how much we expect the sample mean to vary from the true population mean. It is typically derived using the formula:\[ME = z \times SE\]For a 95% confidence interval, the margin of error is approximately 0.2713 days.

Key points about margin of error include:
  • It reflects the maximum expected difference between the sample mean and the population mean within a specified confidence level.
  • The margin of error increases with a larger standard error and decreases with a larger sample size.
  • It is critical to constructing the confidence interval, being added and subtracted from the sample mean \( \bar{x} \) to obtain the interval \[ \bar{x} \pm ME \].
Hence, the margin of error is pivotal in understanding the "range of confidence" for how close our sample estimate is to the population parameter.

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Most popular questions from this chapter

Suppose that \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) denote a random sample of size 4 from a population with an exponential distribution whose density is given by $$f(y)\left\\{\begin{array}{ll} (1 / \theta) e^{-y / \theta}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Let \(X=\sqrt{Y_{1} Y_{2}}\). Find a multiple of \(X\) that is an unbiased estimator for \(\theta\). [Hint: Use your knowledge of the gamma distribution and the fact that \(\Gamma(1 / 2)=\sqrt{\pi}\) to find \(E(\sqrt{Y_{1}})\). Recall that the variables \(Y_{i}\) are independent.] b. Let \(W=\sqrt{Y_{1} Y_{2} Y_{3} Y_{4}}\). Find a multiple of \(W\) that is an unbiased estimator for \(\theta^{2}\). [Recall the hint for part (a).]

In a study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children; in a sample of 100 non-graduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference in the proportions of firstborn or only children for the two populations from which these samples were drawn. Give a bound for the error of estimation.

Suppose that we obtain independent samples of sizes \(n_{1}\) and \(n_{2}\) from two normal populations with equal variances. Use the appropriate pivotal quantity from Section 8.8 to derive a \(100(1-\alpha) \%\) upper confidence bound for \(\mu_{1}-\mu_{2}\)

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Suppose that \(\hat{\theta}\) is an estimator for a parameter \(\theta\) and \(E(\hat{\theta})=a \theta+b\) for some nonzero constants \(a\) and \(b\) a. In terms of \(a, b,\) and \(\theta\), what is \(B(\hat{\theta}) ?\) b. Find a function of \(\hat{\theta}-\operatorname{say}, \hat{\theta}^{*}-\) that is an unbiased estimator for \(\theta\)
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