Answer:
The sample size will be "[tex]102.5494 \sim 102[/tex]". A further explanation is given below.
Step-by-step explanation:
Let the sample size will be:
= n
The given values are:
Standard deviation,
= 1.4
ME,
= 0.3
As we know,
⇒ [tex]n=(Z \frac{a}{2}\times \frac{SD}{E})^2[/tex]
On putting the estimated values, we get
⇒ [tex]=(2.17\times \frac{1.4}{0 .3})^2[/tex]
⇒ [tex]=(\frac{3.038}{0.3})^2[/tex]
⇒ [tex]=102.5494 \sim 102[/tex]
