Question

A random sample of 50 students is chosen from a large population whose diastolic blood pressures has a standard deviation of 5mm Hg. If the 50 students gave a mean pressure of 80 mm Hg, compute the 88.12% confidence interval of the mean of the diastolic pressures of all students.

Answer 1

The following information is provided from 88.12% confidence interval for population mean $\mu$ .

Sample mean $\overline{X}=80$

Sample standard deviation $\:\:\left(s\right)=5\:$

Sample size $\:\left(n\right)=50$

The critical value for $α=0.119$ and $df=n−1=49$ degrees of freedom is $\:\:\:t_c=Z_{1-\frac{\alpha }{2};n-1}=1.588$. The corresponding confidence interval is computed as shown below:

$CI=\left(\overline{X}-t_c\times \frac{s}{\sqrt{n}},\:\overline{X}+t_c\times \:\frac{s}{\sqrt{n}}\:\right)$

$CI\:\:=\left(80-1.588\times \:\:\frac{5}{\sqrt{50}},\:80+1.588\times \:\:\:\frac{5}{\sqrt{50}}\:\right)$

$CI=(78.877, 81.123)$

Therefore, based on the data provided, the 88.12% confidence interval for the population mean is 78.877<μ<81.123, which indicates that we are 88.12% confident that the true population mean μ is contained by the interval (78.877,81.123).