Answer:
The appropriate solution is "95%".
Step-by-step explanation:
The given values are:
Lower bound,
= 173.8
Upper bound,
= 196.2
Mean of cholesterol,
= 138 milligrams
S = 17.6 mg
Now,
The margin of error (E) will be:
= [tex]\frac{Upper \ bound-Lower \ bound}{2}[/tex]
= [tex]\frac{196.2-173.8}{2}[/tex]
= [tex]\frac{22.4}{2}[/tex]
= [tex]11.2[/tex]
As we know,
⇒ [tex]E=t_{\alpha 12}\times \frac{S}{\sqrt{n} }[/tex]
On putting the values, we get
⇒ [tex]11.2=t_{\alpha 12}\times \frac{17.6}{\sqrt{12} }[/tex]
On applying cross-multiplication, we get
⇒ [tex]t_{\alpha 12}=2.2044[/tex]
∴ [tex]\frac{\alpha}{2} =0.025[/tex]
[tex]\alpha = 0.025\times 2[/tex]
[tex]=0.05[/tex]
hence,
The confidence level will be:
⇒ [tex]CI=1-\alpha[/tex]
⇒ [tex]=1-0.05[/tex]
⇒ [tex]=0.95[/tex]
⇒ [tex]=95[/tex]%
