Respuesta :
Answer:
The appropriate response is "(26.03, 33.97)".
Step-by-step explanation:
The given values are:
Average maternal age,
[tex]\bar{x} = 30[/tex]
Standard deviation,
s = 6.2
Random sample,
n = 20
Now,
The degree of freedom will be:
⇒ [tex]df=n-1[/tex]
[tex]=20-1[/tex]
[tex]=19[/tex]
"t" at confidence level 99%, will be:
⇒ [tex]\alpha=1-99 \ percent[/tex]
[tex]=1-0.99[/tex]
[tex]=0.01[/tex]
⇒ [tex]\frac{\alpha}{2}=\frac{0.01}{2}[/tex]
[tex]=0.005[/tex]
⇒ [tex]t\frac{\alpha}{2} \ df=t \ 0.005,19[/tex]
By using the student table, we get
[tex]=2.86 1[/tex]
The margin of error will be:
⇒ [tex]E=\frac{t \alpha}{2,df\times (\frac{s}{\sqrt{n}})}[/tex]
On substituting the estimated values, we get
⇒ [tex]=2.861\times (\frac{6.2}{\sqrt{20}})[/tex]
⇒ [tex]=3.97[/tex]
At 99%, the confidence level will be:
⇒ [tex]\bar{x}-E< \mu <\bar{x}+E[/tex]
⇒ [tex]30-3.97< \mu < 30+3.97[/tex]
⇒ [tex](26.03,33.97)[/tex]
Thus the above is the correct approach.
