Question

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 difference is (-4.664, 1.164).

Step-by-step solution

Identify Given Values

We have two groups: registered nurses and nursing assistants. For the nurses, the sample mean is 30.2, the population standard deviation is 5.6, and the sample size is 30. For the assistants, the sample mean is 31.7, the population standard deviation is 4.3, and the sample size is 40.

Define the Confidence Interval Formula

The formula for the confidence interval of the difference in means is given by: \[ \bar{x}_1 - \bar{x}_2 \pm Z \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \] where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( \sigma_1 \) and \( \sigma_2 \) are the population standard deviations, \( n_1 \) and \( n_2 \) are the sample sizes, and \( Z \) is the z-score for the given confidence level.

Find the Z-Score for 99% Confidence Level

For a 99% confidence level, the Z-score is approximately 2.576.

Plug in Values into the Formula

Use the formula from Step 2: \[ 30.2 - 31.7 \pm 2.576 \times \sqrt{\frac{5.6^2}{30} + \frac{4.3^2}{40}} \] Calculate the difference in sample means, which is \( -1.5 \), and then calculate the standard error.

Calculate the Standard Error

First, compute each fraction: \( \frac{5.6^2}{30} = \frac{31.36}{30} \approx 1.0453 \) \( \frac{4.3^2}{40} = \frac{18.49}{40} \approx 0.46225 \) Then, add them to get the standard error: \( \sqrt{1.0453 + 0.46225} \approx \sqrt{1.50755} \approx 1.228 \).

Compute the Confidence Interval

Multiply the standard error by the Z-score: \( 2.576 \times 1.228 \approx 3.164 \). Find the confidence interval: \[ -1.5 \pm 3.164 \] This gives the range: (-4.664, 1.164).

Interpret the Confidence Interval

The 99% confidence interval for the difference in ages between registered nurses and nursing assistants is from -4.664 to 1.164. This means we are 99% confident that the true difference in mean ages falls within this interval.

Key concepts

Understanding Population Standard Deviation

The population standard deviation, denoted by the symbol \(\sigma\), is a measure of how data points are spread out in a population. It tells us about the typical distance each value in a population is from the mean. In simpler terms, it shows how diverse the ages are within each group.

• Lower standard deviation: Values are closer to the mean.
• Higher standard deviation: Values are more spread out.

For instance, in our original exercise, the population standard deviation for registered nurses' ages is 5.6, while for nursing assistants' ages, it's 4.3. This indicates that the ages of registered nurses are more spread out compared to those of nursing assistants.
By using the population standard deviation in calculations, we ensure that we account for the variability within the entire population, leading to more accurate confidence intervals when comparing groups.

Choosing the Right Sample Size

Sample size, symbolized by \(n\), is the number of observations or data points collected from a population. It plays a crucial role in statistics as it affects the accuracy of our estimations and the margin of error in our results.

• Larger sample sizes tend to provide more reliable results.
• Smaller sample sizes might lead to larger margins of error.

In our nursing example, we have a sample size of 30 registered nurses and 40 nursing assistants. With these sample sizes, we aim to draw conclusions about the entire population of nurses and assistants. Ideally, larger sample sizes reduce the uncertainty (or variability) of the estimates, allowing for a more precise confidence interval calculation. This precision is particularly important when making decisions based on statistical results.

What is a Z-Score and Why It Matters

A z-score is a numerical measurement that describes a value's relation to the mean of a group of values. Z-scores tell us how many standard deviations an element is from the mean. They're crucial in confidence interval calculations, especially in our context where we're dealing with a known population standard deviation.

• A positive z-score signifies a value above the mean.
• A negative z-score indicates a value below the mean.
• A z-score of zero signifies the value is exactly at the mean.

In the exercise, to find the 99% confidence interval, we need a specific z-score, which is approximately 2.576 for 99% confidence. This z-score helps create a range that we're reasonably sure the actual mean difference lies within. This level of confidence reflects our certainty in the estimates given the data.

Interpreting the Difference of Means

The difference of means refers to the difference between the average values of two different groups. In statistics, comparing these means helps determine whether there is a significant difference between the groups and if any observed difference is statistically meaningful or just due to random chance.In the original exercise, the mean age difference between the two groups (registered nurses and nursing assistants) is \(-1.5\), suggesting that on average, registered nurses are younger. By constructing a confidence interval around this mean difference, such as the calculated \[(-4.664, 1.164)\], we can state with 99% confidence that the true difference in mean ages lies within this range.- The negative lower bound indicates the youngest possible mean difference.- The positive upper bound suggests the oldest possible mean difference.
This helps anyone trying to draw conclusions about these groups based on age, while acknowledging some uncertainty in these estimations.