Question
You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 35 bacteria reveals a sample mean of
¯
x
=
66
hours with a standard deviation of
s
=
4.6
hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.75 hours at a 90% level of confidence.
What sample size should you gather to achieve a 0.75 hour margin of error? Round your answer up to the nearest whole number.
¯
x
=
66
hours with a standard deviation of
s
=
4.6
hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.75 hours at a 90% level of confidence.
What sample size should you gather to achieve a 0.75 hour margin of error? Round your answer up to the nearest whole number.
Answers
GPT-4o mini
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.
1. **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 \( Z = 1.645 \).
2. **Substitute the values into the formula**:
\[
n = \left( \frac{1.645 \cdot 4.6}{0.75} \right)^2
\]
3. **Calculate the numerator**:
\[
1.645 \cdot 4.6 \approx 7.577
\]
4. **Calculate the fraction**:
\[
\frac{7.577}{0.75} \approx 10.096
\]
5. **Square the result**:
\[
n \approx (10.096)^2 \approx 102.00
\]
6. **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**.
\[
n = \left( \frac{Z \cdot s}{E} \right)^2
\]
Where:
- \( Z \) is the Z-score corresponding to the desired confidence level.
- \( s \) is the sample standard deviation.
- \( E \) is the margin of error.
1. **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 \( Z = 1.645 \).
2. **Substitute the values into the formula**:
\[
n = \left( \frac{1.645 \cdot 4.6}{0.75} \right)^2
\]
3. **Calculate the numerator**:
\[
1.645 \cdot 4.6 \approx 7.577
\]
4. **Calculate the fraction**:
\[
\frac{7.577}{0.75} \approx 10.096
\]
5. **Square the result**:
\[
n \approx (10.096)^2 \approx 102.00
\]
6. **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**.