A total of 280 observations of Bob​ Ramos, an​ assembly-line worker, were made over a​ 40-hour work week. The sample also showed that Bob was busy working​ (assembling the​ parts) during 250 observations. ​(Round all intermediate calculations to at least two decimal places before proceeding with further​ calculations.)
Part 2
​a) The percent of time Bob is working​ =

Part 3
​b) The number of observations that need to be taken of Bob to be 95​% confident with 5​% acceptable error=

To solve this problem, we'll approach each part separately, using statistical formulas and rounding carefully as required.

### Part 2
#### a) Percent of Time Bob is Working

The percentage of time that Bob is observed working is calculated using the formula:

\[
\text{Percent Working} = \left(\frac{\text{Number of Working Observations}}{\text{Total Observations}}\right) \times 100
\]

Given:
- Total observations = 280
- Working observations = 250

Substituting the values into the formula:

\[
\text{Percent Working} = \left(\frac{250}{280}\right) \times 100 = 89.29\%
\]

Thus, Bob is working approximately **89.29%** of the time.

### Part 3
#### b) Number of Observations Needed for 95% Confidence with 5% Acceptable Error

To determine the number of observations required to estimate the true proportion of time Bob is working, with a 95% confidence level and a 5% margin of error, we use the formula for sample size \( n \):

\[
n = \left(\frac{Z^2 \times p \times (1-p)}{E^2}\right)
\]

Where:
- \( Z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the estimated proportion (from the sample).
- \( E \) is the acceptable margin of error.

Given:
- Confidence level = 95% (so \( Z \approx 1.96 \) for a two-tailed test)
- Estimated proportion \( p = \frac{250}{280} = 0.8929 \)
- Acceptable error \( E = 0.05 \)

Substitute these values into the formula:

\[
n = \left(\frac{1.96^2 \times 0.8929 \times (1 - 0.8929)}{0.05^2}\right)
\]

Let's calculate this value:

\[
n = \left(\frac{3.8416 \times 0.8929 \times 0.1071}{0.0025}\right)
\]

\[
n \approx \frac{0.367 \times 3.8416}{0.0025} \approx \frac{1.4094}{0.0025} \approx 563.76
\]

Rounding up (since you can't take a fraction of an observation), Bob would need approximately **564 observations** to be 95% confident with a 5% margin of error.

---

Would you like more details or have any questions?

### Related Questions

1. How do you determine the z-score for a given confidence level?
2. What is the importance of rounding intermediate calculations?
3. How do sample size calculations change if the margin of error is reduced to 3%?
4. How would the number of observations change for a 99% confidence level?
5. What happens to the required sample size if the proportion observed changes?
6. How can we interpret the margin of error in this context?
7. What are the implications of observing less than the calculated number of observations?
8. How can observational studies be improved for more accuracy in work efficiency?

### Tip
Always round up the calculated sample size to ensure the desired confidence level and margin of error are met.

Learn how to calculate the percentage of time Bob Ramos works and determine the sample size needed for a 95% confidence level with a 5% margin of error. This problem involves statistical concepts including percentage calculation and sample size determination.

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