91Ó°ÊÓ

How much is the cost of a hospital stay increasing? The mean cost of one day in a semiprivate room was reported to be \(\$ 4848\) in 2005 and \(\$ 5260\) in 2006 (The Wall Street Journal, January 2,2007 ). Assume the estimate for 2005 is a sample mean based on a sample size of 80 and the estimate for 2006 is a sample mean based on a sample size of 60 . a. Develop a point estimate of the increase in the cost of a semiprivate hospital room from 2005 to 2006 b. Historical data indicate that a population standard deviation of \(\$ 800\) is a reasonable assumption for both years. Compute the margin of error for your estimate in part (a). Use \(95 \%\) confidence. c. Develop a \(95 \%\) confidence interval estimate of the increase in cost for a semiprivate room.

Step by step solution

01

Calculate the Point Estimate

The point estimate for the increase in cost from 2005 to 2006 is calculated by subtracting the mean cost in 2005 from the mean cost in 2006.\[\text{Point Estimate} = \bar{x}_2 - \bar{x}_1 = 5260 - 4848 = 412\]

02

Find the Standard Error of the Difference in Means

The standard error of the difference between the two sample means is calculated using the formula:\[SE = \sqrt{\left(\frac{\sigma^2}{n_1}\right) + \left(\frac{\sigma^2}{n_2}\right)}\]Given that \(\sigma = 800\), \(n_1 = 80\), and \(n_2 = 60\), we have:\[SE = \sqrt{\frac{800^2}{80} + \frac{800^2}{60}} = \sqrt{8000 + 10666.67} = \sqrt{18666.67} \approx 136.61\]

03

Calculate the Margin of Error

The margin of error (ME) for a 95% confidence interval is given by:\[ME = z_{0.025} \times SE\]where \(z_{0.025} \approx 1.96\) for a 95% confidence level. Substituting the values:\[ME = 1.96 \times 136.61 \approx 267.75\]

04

Develop the Confidence Interval

Using the point estimate and the margin of error, the 95% confidence interval for the increase in cost is:\[(\bar{x}_2 - \bar{x}_1 - ME, \bar{x}_2 - \bar{x}_1 + ME) = (412 - 267.75, 412 + 267.75) = (144.25, 679.75)\]

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Key Concepts

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

Point Estimate

When working with statistical data, oftentimes we want to make informed guesses about a population parameter. A **point estimate** serves as our best single guess about that parameter.
In the context of our exercise, we calculated how much the average cost of a hospital stay increased from 2005 to 2006. We did this by finding the difference between the mean costs of the two years.
This is a straightforward calculation:

• The cost in 2005 was reported at \(\\(4848\).
• The cost in 2006 was \(\\)5260\).
• Thus, the point estimate for the increase is \(\\(5260 - \\)4848 = \$412\).

This point estimate is crucial as it gives us a starting value to work with, representing the most probable increase based on our sample data.

Standard Error

To understand the precision of our point estimate, we look at the **standard error** (SE). It helps measure how much our sample mean (point estimate) might fluctuate if we were to take multiple samples.
In simpler terms, SE tells us about the potential error in our estimation.
For comparing the increase between 2005 and 2006, we calculate the SE using the formula:

• \[SE = \sqrt{\left(\frac{\sigma^2}{n_1}\right) + \left(\frac{\sigma^2}{n_2}\right)}\]
• Where \(\sigma\) is the population standard deviation (\(\$800\)), \(n_1\) and \(n_2\) are sample sizes for 2005 and 2006 respectively.
• The calculation gives us an SE of approximately \(136.61\).

This SE suggests how much variability we might expect in the increase estimate if we looked at different samples of the same size. Lower SE means more precision in our estimate.

Margin of Error

The **margin of error** (ME) represents the range within which we expect the true difference (in our case, the increase in room cost) to lie, with a certain level of confidence.
It combines the standard error with a critical value from the normal distribution (like a \(z\) score).
For a 95% confidence interval, the formula is:

• \[ME = z_{0.025} \times SE\]
• The critical value \(z_{0.025}\) equals approximately \(1.96\).
• Thus, the margin of error for our exercise is: \(1.96 \times 136.61 \approx 267.75\).

This means we are 95% confident that the true increase in room cost is \(\pm 267.75\) around our point estimate. By combining the point estimate and margin of error, we can create a confidence interval, offering more comprehensive insights into the data.