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 90% confidence interval to estimate the mean vitamin D level in the population = (26.22,27.78)

Interpretation: We are 90% confident that the true population mean lies in  (26.22,27.78).

Step-by-step explanation:

Let X denotes a random variable that represents the vitamin D level .

As per given,

Since population standard deviation is unknown , so we will use t -test .

For that, Degree of freedom = df = n-1 = 41

Significance

Two-tailed critical t-value :

Confidence interval for population mean:

So, a 90% confidence interval to estimate the mean vitamin D level in the population = (26.22,27.78)

Interpretation: We are 90% confident that the true population mean lies in  (26.22,27.78).

15.36 per mile

but forgot to round

i am pretty sure