Question

Age Differences In a large hospital, a nursing director selected a random sample of 30 registered nurses and found that the mean of their ages was \(30.2 .\) The population standard deviation for the ages is \(5.6 .\) She selected a random sample of 40 nursing assistants and found the mean of their ages was \(31.7 .\) The population standard deviation of the ages for the assistants is \(4.3 .\) Find the \(99 \%\) confidence interval of the differences in the ages.

Short answer

The 99% confidence interval for the age difference is [-4.664, 1.164].

Step-by-step solution

Identify the Variables

For the registered nurses, we have a sample mean \( \bar{x_1} = 30.2 \), sample size \( n_1 = 30 \), and population standard deviation \( \sigma_1 = 5.6 \). For the nursing assistants, we have a sample mean \( \bar{x_2} = 31.7 \), sample size \( n_2 = 40 \), and population standard deviation \( \sigma_2 = 4.3 \). We aim to find the confidence interval for the difference in means \( \mu_1 - \mu_2 \).

Define the Confidence Interval Formula

For confidence intervals for the difference in means of two independent samples with known population standard deviations, we use:\[(\bar{x_1} - \bar{x_2}) \pm Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]Since \( 99\% \) confidence interval is required, find \( Z_{\alpha/2} \) for \( \alpha = 0.01 \).

Find the Z-value

Use a standard normal distribution table to find \( Z_{\alpha/2} \) for a \( 99\% \) confidence level, which gives \( 0.5 - 0.495 = 0.005 \) in each tail. This corresponds to \( Z_{0.005} = 2.576 \).

Calculate the Standard Error

Calculate the standard error for the difference in means:\[SE = \sqrt{\frac{5.6^2}{30} + \frac{4.3^2}{40}} = \sqrt{1.04533 + 0.46225} = \sqrt{1.50758} \approx 1.228\]

Calculate the Confidence Interval

Substitute values into the confidence interval formula:\[(30.2 - 31.7) \pm 2.576 \times 1.228\]\[= -1.5 \pm 3.164\]This gives the interval:\[[-4.664, 1.164]\]

Interpret the Confidence Interval

The \( 99\% \) confidence interval for the difference in ages between registered nurses and nursing assistants is \([-4.664, 1.164]\). This means we are 99% confident that the true difference in average ages lies within this range.

Key concepts

Difference in Means

When we talk about the **difference in means**, we're comparing the average values from two different groups. In our case, we have registered nurses and nursing assistants. Their ages have been measured, and we want to see how their average ages differ.
But it's not just about subtracting one mean from the other. We use specific statistical techniques to ensure our results are reliable and accurate.

• To calculate the difference in means, we start with the mean age of registered nurses ( \( \bar{x_1} = 30.2 \) years ).
• Then, we subtract the mean age of nursing assistants ( \( \bar{x_2} = 31.7 \) years ) from this value.

This gives us: \( \bar{x_1} - \bar{x_2} = 30.2 - 31.7 = -1.5 \)
This value reflects a negative difference, suggesting that, on average, registered nurses are younger than nursing assistants.

Population Standard Deviation

The **population standard deviation** is a measure of the spread of ages within each group. It tells us how much individual ages in each group deviate from the average.
Unlike a sample standard deviation, which is calculated from just the sample data, the population standard deviation is typically known or assumed because it encompasses the whole population's data.

• For registered nurses, the population standard deviation is given as \( \sigma_1 = 5.6 \).
• For nursing assistants, the population standard deviation is \( \sigma_2 = 4.3 \).

In terms of interpreting these numbers, a smaller standard deviation indicates that the ages of nursing assistants are more closely clustered around the mean compared to the ages of registered nurses.

Standard Error

The **standard error** (SE) is a crucial part of estimating how the sample mean (or the difference between two sample means) represents the population. It allows us to understand how much the means we've calculated could vary just by chance alone.
Specifically, when dealing with two independent samples, the standard error of the difference in means is calculated using the formula: \[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]

• In our example, \( \sigma_1 = 5.6 \), \( \sigma_2 = 4.3 \), \( n_1 = 30 \), \( n_2 = 40 \).
• Plugging in these values, we find the standard error to be approximately \( SE \approx 1.228 \).

This SE value tells us about the reliability of our estimated difference. Smaller SE means a more precise estimate.

Z-value

An understanding of the **Z-value** is crucial when working with confidence intervals, as it provides the critical value needed to estimate where our true population mean lies.
In a standard normal distribution - or a bell curve with a mean of zero and a standard deviation of one - Z-values indicate how many standard deviations an element is from the mean.

• For a 99% confidence interval, we seek a Z-value that puts 0.5% of data in each of the curve’s tails.
• This corresponds to a Z-value of approximately 2.576, obtained from standard Z-tables or calculators.

By using this Z-value along with our standard error, we can build a confidence interval. This interval estimates the range within which the true mean difference likely falls, factoring in potential variations due to random sample selection.