Answer:
Sample size n [tex]\simeq[/tex] 1269.15
Step-by-step explanation:
From the information given ,
At 99% of confidence interval,
the level of significance ∝ = 1 - 0.99
the level of significance ∝ = 0.01
the critical value for 99% of confidence interval is:
[tex]\mathtt{\dfrac{\alpha }{2} = \dfrac{0.01}{2}}[/tex]
= 0.005
[tex]\mathtt {z_{\alpha/2} = z_{0.005/2} }[/tex]
The value for z from the standard normal tables
= 2.58
The Margin of error E= 3% = 0.03
The formula to determine the sample size n used can be expressed as follows:
[tex]\mathtt { n = (\dfrac{z_{\alpha/2}}{E})^2 \ \hat p (1 - \hat p) }[/tex]
where;
[tex]\mathtt{\hat p }[/tex] = 22% = 0.22
Then:
[tex]\mathtt { n = (\dfrac{2.58}{0.03})^2 \ \times 0.22 \times (1 - 0.22) }[/tex]
[tex]\mathtt { n = (86)^2 \ \times 0.22 \times (0.78) }[/tex]
[tex]\mathtt { n = 7396 \ \times 0.22 \times (0.78) }[/tex]
n = 1269.1536
Sample size n [tex]\simeq[/tex] 1269.15
