Question

Serum iron (mmol/L) was measure in 60 healthy adults. The sample mean serum iron was 18.30 (mmol/L) with a sample standard deviation of 4.66 (mmol/L) a. Based on these data estimate the population mean serum iron level in healthy adults. b. Based on these data estimate the standard error of the sample mean.

Answer 1

To estimate the population mean serum iron level, use a confidence interval with the sample mean and standard deviation. The population mean is estimated to be between 15.44 and 21.16 (mmol/L). The standard error of the sample mean is 0.60 (mmol/L).

Step-by-step explanation:

To estimate the population mean serum iron level in healthy adults based on the sample mean and standard deviation, we can use a confidence interval. We'll assume that the distribution of serum iron levels is approximately normal. A 95% confidence interval can be calculated using the formula:

CI = sample mean +/- (critical value * (sample standard deviation / square root of sample size))

For a 90% confidence interval, the critical value is 1.645.

So, the population mean serum iron level is estimated to be between 15.44 (18.30 - (1.645 * (4.66 / square root of 60))) and 21.16 (18.30 + (1.645 * (4.66 / square root of 60))).

To estimate the standard error of the sample mean, we divide the sample standard deviation by the square root of the sample size:

Standard error = sample standard deviation / square root of sample size

Therefore, the standard error of the sample mean is 0.60 (4.66 / square root of 60).