Answer: 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,
[tex]\text{Sample size=}n=42\\\\ \text{Sample mean=}\overline{x}=27\\\\ \text{Sample standard deviation}= s= 3[/tex]
Since population standard deviation is unknown , so we will use t -test .
For that, Degree of freedom = df = n-1 = 41
Significance [tex]\alpha=1-0.90=0.1[/tex]
Two-tailed critical t-value : [tex]t_{\alpha/2,n-1}=t_{0.05,41}=1.683[/tex]
Confidence interval for population mean:
[tex]\overline{x}\pm t_{\alpha/2,n-1}\dfrac{s}{\sqrt{n}}\\\\ =27\pm (1.683)\dfrac{3}{\sqrt{42}}\\\\ =27\pm (1.683)\dfrac{3}{6.4807}\\\\=27\pm (1.683)(0.463)\\\\=27\pm0.78\\\\=(27-0.78,\ 27+0.78)\\\\=(26.22,27.78)[/tex]
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).
