To determine the required sample size \( n \) to estimate the mean lifespan of the bacteria within a margin of error \( E \) at a given confidence level, we can use the formula for the sample size in the context of estimating a population mean:
\[ n = \left( \frac{Z \cdot s}{E} \right)^2 \]
Where:
- \( Z \) is the Z-score corresponding to the desired confidence level.
- \( s \) is the sample standard deviation.
- \( E \) is the margin of error.
-
Identify the parameters:
- Margin of error \( E = 0.75 \) hours.
- Standard deviation \( s = 4.6 \) hours.
- Confidence level = 90%. The Z-score for a 90% confidence level is approximately \( Z = 1.645 \).
-
Substitute the values into the formula: \[ n = \left( \frac{1.645 \cdot 4.6}{0.75} \right)^2 \]
-
Calculate the numerator: \[ 1.645 \cdot 4.6 \approx 7.577 \]
-
Calculate the fraction: \[ \frac{7.577}{0.75} \approx 10.096 \]
-
Square the result: \[ n \approx (10.096)^2 \approx 102.00 \]
-
Round up to the nearest whole number: \[ n = 103 \]
Thus, the required sample size to achieve a margin of error of 0.75 hours at a 90% confidence level is 103.