A researcher is interested in assessing the blood folate level of women who live in a population with a high incidence of neural tube defects. She takes a large random sample (n) of women age 18-45 and measures their blood folate levels. The researcher finds that the data are normally distributed, and she reports the mean and standard deviation (SD) of the sample. To account for sampling variation, she decides to calculate a 95% confidence interval to estimate the mean of the entire population. The researcher concludes that 2.4 to 4.6 ng/mL might be a likely range for the true, unknown population mean. Which of the following calculations was most likely used to compute this interval estimate of the population mean?
The normal distribution is a continuous distribution that follows a bell-shaped curve with a single peak at the mean value. One of the most important properties of the normal distribution is its symmetry about its mean, which allows for determination of the percentage of all observations that lie within 1, 2, or 3 standard deviations (SDs). The "68/95/99 rule" states that 68% of all observations lie within 1 SD of the mean, 95% within 2 SDs of the mean, and 99.7% within 3 SDs of the mean. These are helpful approximations; however, when considering the normal distribution, exactly 95% of the observations lie within 1.96 SDs of the mean and 99% of the observations lie within 2.58 SDs (1.96 and 2.58 represent z-scores for 95% and 99% of the distribution, respectively).
Most research is done using samples rather than entire populations. This introduces some variability when calculating population parameters such as mean value (eg, when different samples are drawn from the same population, the means from all possible samples will often be slightly different). The variability between sample means can be calculated in a way analogous to calculating the SD of a group of observations in a single sample. The "standard deviation" of a series of sample means is known as the standard error (SE) of the mean, and it estimates how far the sample mean is likely to be from the unknown population mean. The SE is estimated considering both the SD and the size of the sample (n) in the following manner: SD/√n.
A sample mean is a point estimate of the true population mean; however, a confidence interval (CI) better accounts for the variability due to sampling by including the SE in its calculation. The CI of the mean can be calculated as follows:
CI of mean = mean ± [z-score for confidence level] × [SE]
where the z-score for confidence level represents the number of SEs containing the desired percentage of observations around the mean (eg, 95%, 99%). In this case, the 95% CI can be calculated:
CI of mean = mean ± 1.96 × (SD/√n)
The sample size and the SD of the sample determine the magnitude of the variability due to sampling. As n increases, SE decreases, and the CI becomes narrower and more precise. Conversely, as the SD increases, SE increases, and the CI becomes wider and less precise.
(Choice A) CI estimates use the SE of the mean, which takes into account the SD and the size of the sample (n). As such, the equation divides the SD by the square root of n.
(Choices C, D, and E) As noted, the SD reflects the spread of individual values in a normal distribution. Multiplying the SD by a specific constant (z-score) gives us a range of values that encompass a certain proportion of the observations. The mean ± 1 × SD would cover 68% of the observations, and the mean ± 2.58 × SD would cover 99% of the observations. However, the question is asking for a 95% CI, so the equation should use a z-score of 1.96 and the SE (ie, SD/√n).
Educational objective:
The standard deviation reflects the spread of individual values in a normal distribution (ie, it measures the variability of the observations within a single sample). The standard error of the mean reflects the variability of means (ie, variance between the means of different samples) and helps estimate the true mean of the underlying population.