genesisdiaz5123
25.09.2020 •
Mathematics
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.
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Ответ:
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).
Ответ:
B. (1, -1)
C. (5, -1)
D. (5, -3)
I hope this helps