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

Answer:

$8.636-1.653\frac{3.9265}{\sqrt{187}}=8.1614$    

$8.636+1.653\frac{3.9265}{\sqrt{187}}=9.1106$    

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

Step-by-step 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:

$8.636-1.653\frac{3.9265}{\sqrt{187}}=8.1614$    

$8.636+1.653\frac{3.9265}{\sqrt{187}}=9.1106$    

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