To determine the required sample size for estimating the prevalence of malnutrition with given parameters:
1. Understand the Given Parameters:
- Prevalence rate of malnutrition ([tex]\( p \)[/tex]): 50% or 0.50
- Complement of the prevalence rate ([tex]\( q \)[/tex]): [tex]\( 1 - p \)[/tex] which is 50% or 0.50
- Margin of error ([tex]\( e \)[/tex]): 8% or 0.08
- Confidence level: 95%
- Non-response rate: 5% or 0.05
- Z-score for 95% confidence level ([tex]\( z \)[/tex]): 1.96 (this value is derived from the standard normal distribution table)
2. Calculate Initial Sample Size ([tex]\( n_0 \)[/tex]):
- Formula to calculate the initial sample size without accounting for non-response rate:
[tex]\[
n_0 = \frac{z^2 \cdot p \cdot q}{e^2}
\][/tex]
- Plugging in the values, we have:
[tex]\[
n_0 = \frac{(1.96)^2 \cdot 0.50 \cdot 0.50}{(0.08)^2}
\][/tex]
- Performing the calculation:
[tex]\[
n_0 = \frac{3.8416 \cdot 0.25}{0.0064} = \frac{0.9604}{0.0064} \approx 150.0625
\][/tex]
Therefore, the initial sample size [tex]\( n_0 \)[/tex] is approximately [tex]\( 150.06 \)[/tex].
3. Adjust for Non-response Rate:
- To account for a non-response rate of 5%, we need to adjust the initial sample size.
- The final sample size is calculated using the formula:
[tex]\[
\text{Final Sample Size} = \frac{n_0}{1 - \text{Non-response rate}}
\][/tex]
- Plugging in the values:
[tex]\[
\text{Final Sample Size} = \frac{150.0625}{1 - 0.05} = \frac{150.0625}{0.95} \approx 157.96
\][/tex]
Therefore, the final sample size accounting for the non-response rate is approximately [tex]\( 157.96 \)[/tex].
Summary:
- The initial sample size ([tex]\( n_0 \)[/tex]) without accounting for the non-response rate is approximately 150.06.
- After adjusting for a non-response rate of 5%, the final sample size needed is approximately 157.96.
