A federal report indicated that 26% of children under age 6 live in poverty in Oklahoma, an increase over previous years. How large a sample is needed to estimate the true proportion of children under age 6 living in poverty in Oklahoma within 2% with 99% confidence? Round the intermediate calculations to three decimal places and round up your final answer to the next whole number.

Answer:

Given:

The population proportion $(p)=0.26$

The margin of error $(e)=0.02$

The confidence interval level $=99%$

Solution:

→ The critical value at 99% confidence interval = 2.576

→ The sample size $(n):$

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

$$\therefore e = z_c \cdot \sqrt{\frac{p(1 – p)}{n}} \quad $$ $$ \therefore 0.02 = 2.576 \cdot \sqrt{\frac{0.26(1 – 0.26)}{n}} \quad $$ $$ \therefore n = 3{,}192$$

Final Answer:
The sample size $(n) = 3192$