contestada

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.)

Respuesta :

Answer:

[tex]8.636-1.653\frac{3.9265}{\sqrt{187}}=8.1614[/tex]    

[tex]8.636+1.653\frac{3.9265}{\sqrt{187}}=9.1106[/tex]    

And we are confident that the true mean for this case is given by [tex]8.1614 \leq \mu \leq 9.1106[/tex]

Step-by-step explanation:

Data given

[tex]\bar X=8.636[/tex] represent the sample mean for the WBC

[tex]\mu[/tex] 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:

[tex]df=n-1=187-1=186[/tex]

The Confidence level is 0.90 or 90%, the significance is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], the critical value for this case would be [tex]t_{\alpha/2}=1.653[/tex]

Replacng into the formula we got:

[tex]8.636-1.653\frac{3.9265}{\sqrt{187}}=8.1614[/tex]    

[tex]8.636+1.653\frac{3.9265}{\sqrt{187}}=9.1106[/tex]    

And we are confident that the true mean for this case is given by [tex]8.1614 \leq \mu \leq 9.1106[/tex]