Answer :
1. 1285 samples for 98% confidence, 3% margin, with 0.33 proportion.
2. 3886 samples for 95% confidence, 1.5% margin, with 0.24 proportion.
3. 243 samples for 98% confidence, 10 margin, with 66.9 standard deviation.
To find the required sample sizes for estimating population proportions and means with specified confidence levels and margins of error, we can use the formulas for sample size calculation:
For estimating population proportions:
[tex]\[ n = \frac{z^2 \cdot p \cdot (1-p)}{E^2} \][/tex]
For estimating population means:
[tex]\[ n = \frac{z^2 \cdot \sigma^2}{E^2} \][/tex]
Where:
- ( n ) is the required sample size
- ( z ) is the z-score corresponding to the desired confidence level (e.g., 98% or 95%)
- ( p ) is the estimated population proportion (for proportion estimation)
[tex]- \( \sigma[/tex] ) is the estimated population standard deviation (for mean estimation)
- E ) is the desired margin of error
Let's calculate the sample sizes for each scenario:
1. Estimating a population proportion with previous evidence suggesting ( p = 0.33 ), desired confidence level ( 98%), and margin of error ( 3%):
z = 2.33 (from standard normal distribution for 98% confidence level)
[tex]\[ n = \frac{(2.33)^2 \cdot 0.33 \cdot (1-0.33)}{(0.03)^2} \][/tex]
[tex]\[ n \approx \frac{5.4289 \cdot 0.33 \cdot 0.67}{0.0009} \]\[ n \approx \frac{1.1515}{0.0009} \]\[ n \approx 1285 \][/tex]
So, a sample size of approximately 1285 is required.
2. Estimating a population proportion with previous evidence suggesting p = 0.24 ), desired confidence level [tex]\( 95\% \)[/tex] , and margin of error [tex]\( 1.5\% \):[/tex]
z = 1.96 (from standard normal distribution for 95% confidence level)
[tex]\[ n = \frac{(1.96)^2 \cdot 0.24 \cdot (1-0.24)}{(0.015)^2} \]\[ n \approx \frac{3.8416 \cdot 0.24 \cdot 0.76}{0.000225} \]\[ n \approx \frac{0.876}{0.000225} \]\[ n \approx 3885.333 \][/tex]
So, a sample size of approximately 3886 is required.
3. Estimating a population mean with previous evidence suggesting
( sigma = 66.9 ), desired confidence level [tex]\( 98\% \),[/tex] and margin of error ( 10 ):
z = 2.33 (from standard normal distribution for 98% confidence level)
[tex]\[ n = \frac{(2.33)^2 \cdot (66.9)^2}{(10)^2} \]\[ n \approx \frac{5.4289 \cdot 4467.61}{100} \]\[ n \approx \frac{24275.5129}{100} \]\[ n \approx 242.755 \][/tex]
So, a sample size of approximately 243 is required.
