Answer:
The appropriate answer is "523".
Step-by-step explanation:
Given:
Margin of error,
E = 0.04
Confidence level,
= 95%
Educated guess,
[tex]P_g[/tex] = 0.32
According to the question,
[tex]\alpha = \frac{100-95}{100}[/tex]
[tex]=0.05[/tex]
[tex]\frac{\alpha}{2} = \frac{0.05}{2}[/tex]
[tex]=0.025[/tex]
[tex]Z_{0.025} = 1.96[/tex]
The sample size will be:
⇒ [tex]n=P_g (1-P_g) (\frac{Z_{\frac{\alpha}{2} }}{E} )^2[/tex]
By substituting the values, we get
[tex]=0.32(1-0.32)(\frac{1.96}{0.04} )^2[/tex]
[tex]=0.32\times 0.68\times (49)^2[/tex]
[tex]=0.32\times 0.68\times 2401[/tex]
[tex]=522.4576[/tex]
or,
[tex]=523[/tex]
