To solve the problem:

Formula:

The formula to calculate the required sample size ($n$) for estimating the mean is:

$n = \left(\frac{z \cdot \sigma}{E}\right)^2$

Where:

• $z$ = z-score corresponding to the desired confidence level (for 95% confidence, $z = 1.96$).
• $\sigma = 17.3$ (the standard deviation provided).
• $E = 1$ (the margin of error given as "within one book").

Calculation:

1. Plug in the values: $n = \left(\frac{1.96 \cdot 17.3}{1}\right)^2$

2. Simplify: $n = \left(33.908\right)^2$

3. Square the value: $n \approx 1149.10$

4. Round up to the nearest whole number: $n = 1150$

Answer:

A 95% confidence level requires 1150 subjects.

Let me know if you want a detailed breakdown of the steps or have further questions!

Related Questions:

1. What is the relationship between margin of error and sample size?
2. How does increasing the confidence level affect the required sample size?
3. What happens if the standard deviation is reduced in this calculation?
4. How would the z-score change for a 99% confidence level?
5. What if the margin of error were set to 2 books instead of 1?

Tip:

Always round up the sample size calculation to ensure the desired confidence level and margin of error are met!