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Mathematics, 12.08.2020 05:01 marjorieheckaman

Vitamin D (delivered naturally by sunshine) helps the human body stay energized and bum fat. Researchers sampled 42 random sunbathers. The average vitamin D level was 27 nanograms per milliliter and the standard deviation was 3 nanograms per militer. 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). 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,
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).
The 90% confidence interval is (26.221, 27.779). Ninety percent of all samples of this size will yield a confidence interval ot (26.221, 27.779).
O 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.
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.

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Vitamin D (delivered naturally by sunshine) helps the human body stay energized and bum fat. Researc...
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