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. The 90% confidence interval is (25.7496, 28.2504). A. 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 (25.7496, 28.2504). Ninety percent of all samples of this size will yield a confidence interval of (25.7496, 28.2504).
C. 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).
D. 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.
E. The 90% confidence interval is (26.221, 27.779). We are 90% confident that the true population mean of vitamin D levels for sunbathers will be between 26.221 nanograms per milliliter and 27.779 nanograms per milliliter.