Question

Vitamin D (delivered naturally by sunshine) helps the human body stay energized and burn fat. Researchers sampled 42 random sunbathers. The average vitamin D level was 27 nanograms per milliliter and the standard deviation was 3 nanograms per milliliter. Construct and interpret a 90% confidence interval to estimate the mean vitamin D level in the population.

a) The 90% confidence interval is (25.7496, 28.2504). We are 90% confident that the true population mean level of vitamin D in sunbathers will be between 25.7496 nanograms per milliliter and 28.2504 nanograms per milliliter.
b) The 90% confidence interval is (26.221, 27.779). Ninety percent of all samples of this size will yield a confidence interval of (26.221, 27.779).
c) The 90% confidence interval is (26.221, 27.779). There is a 90% chance that a randomly selected sunbather is one whose vitamin D level lies between 26.22 nanograms per milliliter and 27.779 nanograms per milliliter.
d) The 90% confidence interval is (28.2504, 25.7496). Ninety percent of all samples of this size will yield a confidence interval of (28.2504, 25.7496).

Answer 1

To construct a 90% confidence interval to estimate the mean vitamin D level in the population, we use the formula: Confidence interval = sample mean ± (critical value x standard deviation / square root of sample size). In this case, the sample mean is 27 nanograms per milliliter and the standard deviation is 3 nanograms per milliliter. The confidence interval is approximately (25.7496, 28.2504).

Step-by-step explanation:

To construct a 90% confidence interval to estimate the mean vitamin D level in the population, we use the formula:

Confidence interval = sample mean ± (critical value x standard deviation / square root of sample size)

In this case, the sample mean is 27 nanograms per milliliter and the standard deviation is 3 nanograms per milliliter. The critical value for a 90% confidence level with a sample size of 42 can be found using a t-distribution table and is approximately 1.684.

Plugging in these values into the formula, we get:

Confidence interval = 27 ± (1.684 x 3 / √42)

Simplifying the expression, the confidence interval is approximately (25.7496, 28.2504). Therefore, we can conclude that we are 90% confident that the true population mean level of vitamin D in sunbathers will be between 25.7496 nanograms per milliliter and 28.2504 nanograms per milliliter.