Question

Twenty-five blood samples were selected by taking every seventh blood sample from racks holding 187 blood samples from the morning draw at a medical center. The white blood count (WBC) was measured using a Coulter Counter Model S. The mean WBC was 8.636 with a standard deviation of 3.9265. (a) Construct a 90% confidence interval for the true mean using the FPCF. (Round your answers to 4 decimal places.)

Answer 1

`8.636-1.653(3.9265)/(√(187))=8.1614`

`8.636+1.653(3.9265)/(√(187))=9.1106`

And we are confident that the true mean for this case is given by
`8.1614 \leq \mu \leq 9.1106`

Explanation:

Data given

`\bar X=8.636` represent the sample mean for the WBC

`\mu` population mean

s=3.9265 represent the sample standard deviation

n=187 represent the sample size

Confidence interval

The confidence interval for the true mean is given by:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

The degrees of freedom, are given by:

`df=n-1=187-1=186`

The Confidence level is 0.90 or 90%, the significance is
`\alpha=0.1` and
`\alpha/2 =0.05`, the critical value for this case would be
`t_(\alpha/2)=1.653`

Replacng into the formula we got:

`8.636-1.653(3.9265)/(√(187))=8.1614`

`8.636+1.653(3.9265)/(√(187))=9.1106`

And we are confident that the true mean for this case is given by
`8.1614 \leq \mu \leq 9.1106`