The estimate of the population proportion is to be within plus or minus 0.04, with an 84% level of confidence. How large a sample is required? (The final answer must be a whole number.)

Answer:

Given Data:

The margin of error $(e)=0.04$

The confidence interval level $=84%$

Here the sample proportion is not given so we can use $(p)=0.5$

Solution:

The level of significance $(\alpha)=1-0.84=0.16$

The critical value $(z_c)=Z_{\frac{\alpha}{2}} = Z_{\frac{0.16}{2}} = 1.41 $

Here we need to use a margin of error formula to find out the sample size:

$$\therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}}$$ $$\therefore 0.04 = 1.41 \cdot \sqrt{\frac{0.5(1 – 0.5)}{n}} $$ $$\therefore n = 311$$