Question

A group of 101 patients with end-stage renal disease were given the drug epoetin. \({ }^{19}\) The mean hemoglobin level of the patients was \(10.3(\mathrm{~g} / \mathrm{dl})\), with an \(\mathrm{SD}\) of 0.9 . Construct a \(95 \%\) confidence interval for the population mean.

Short answer

The 95% confidence interval for the population mean is approximately (10.12 g/dl, 10.48 g/dl).

Step-by-step solution

Determine the formula for the confidence interval

The confidence interval for the population mean, when the population standard deviation is unknown and the sample size is relatively small, can be calculated using the formula \[\bar{X} \pm t_{\frac{\alpha}{2}} \left(\frac{s}{\sqrt{n}}\right)\], where \[\bar{X}\] is the sample mean, \[t_{\frac{\alpha}{2}}\] is the t-score for \(95\%\) confidence interval with \(n-1\) degrees of freedom, \[s\] is the sample standard deviation, and \[n\] is the sample size.

Identify the variables from the given data

In the given exercise, the sample size \(n\) is 101, the sample mean \(\bar{X}\) is 10.3 g/dl, and the sample standard deviation \(s\) is 0.9 g/dl. Since the sample size is greater than 30, it would also be reasonable to use the z-score. However, we'll use the t-score because the problem doesn't specify whether to use z-score or t-score, and the t-distribution is more accurate for smaller sample sizes. We are constructing a \(95\%\) confidence interval, so we'll need to find the t-score that corresponds to \(95\%\) confidence and 100 degrees of freedom.

Find the appropriate t-score

To find the t-score \(t_{\frac{\alpha}{2}}\) for a \(95\%\) confidence interval and 100 degrees of freedom, you can use a t-distribution table or an online calculator. The value is approximately 1.984 for \(95\%\) confidence and 100 degrees of freedom.

Calculate the margin of error

The margin of error (ME) is calculated as \[ME = t_{\frac{\alpha}{2}} \left(\frac{s}{\sqrt{n}}\right)\]. Substituting the values we get \[ME = 1.984 \left(\frac{0.9}{\sqrt{101}}\right)\]. Calculating this gives us the margin of error to add and subtract from the sample mean.

Construct the confidence interval

Finally, we use the margin of error to find the lower and upper limits of the confidence interval as follows: \[\text{Lower Limit} = \bar{X} - ME\] \[\text{Upper Limit} = \bar{X} + ME\] Substitution of the sample mean and the margin of error into these equations gives the bounds of the \(95\%\) confidence interval for the population mean.

Key concepts

T-Score

The t-score is a statistical measure that shows how many standard deviations a data point is from the sample mean. It’s especially useful when dealing with small sample sizes or when the population standard deviation is unknown. A key use of the t-score is in constructing confidence intervals.

In the context of our exercise, the t-score helps us determine how confident we can be that the population mean falls within a certain range. For a 95% confidence interval with 100 degrees of freedom, we used a t-score of approximately 1.984, which was found using a t-table or an online calculator. This t-score reflects a point on a t-distribution curve which has 95% of the data points to its left, ensuring that our interval is capturing the central 95% portion of the possible means.

Population Mean

The population mean is the average of all measurements in the entire population. It is an essential concept in statistics, representing the central value of a data set. However, most of the time, the population mean is not known, and we must estimate it using sample data.

In our exercise, we are trying to estimate this value for patients with end-stage renal disease using the hemoglobin levels from a sample of 101 patients taking epoetin. The goal of the confidence interval is to give us a range within which we expect the population mean to lie, with a given level of certainty (95% in our case).

Sample Standard Deviation

Sample standard deviation (s) measures the dispersion of sample data points from the sample mean. It differs from population standard deviation in that it uses sample data instead of the entire population data and uses n-1 (degrees of freedom) instead of n when dividing the sum of squared differences from the mean. The rationale behind this is to compensate for the fact that the sample mean is an estimate of the population mean and generally leads to a less biased estimate of the population standard deviation, especially for smaller samples.

For the patients in our exercise, a sample standard deviation of 0.9 g/dl means there’s a certain amount of variability in the hemoglobin levels around the sample mean of 10.3 g/dl.

Margin of Error

The margin of error measures the extent of the uncertainty in our estimate of the population mean. It tells us how much we can expect the sample mean to vary from the population mean. The margin of error is affected by the level of confidence we desire (e.g., 95% versus 99%) and the variability of the data, as measured by the standard deviation.

In our problem, we found a margin of error by multiplying the t-score (which corresponds to our confidence level) by the sample standard deviation divided by the square root of the sample size. This margin of error is then used to construct a range around the sample mean, which gives us our confidence interval.

Degrees of Freedom

Degrees of freedom (df) in statistics represent the number of values in a calculation that are free to vary. When we calculate sample standard deviation, we use n-1 degrees of freedom, where n is the sample size. This is because we are estimating the population standard deviation from the sample, and one value (the sample mean) is already fixed, limiting the variance in the remaining data points.

In the exercise example, the degrees of freedom are 100 (101 - 1), which reflects that we have 100 independent pieces of information available to estimate the population parameter. The degrees of freedom are crucial in determining the correct t-score for our confidence interval calculation.