An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous​ studies, the standard deviation is assumed to be 21 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within ± 5 minutes.

**a.** What sample size is​ needed?

**a.** The sample size required for​ 99% confidence is

enter your response here

​(Round up to the nearest​ integer.)

To determine the required sample size for estimating the mean with a certain level of confidence and margin of error, we can use the following formula:

\[
n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2
\]

Where:
- \( Z_{\alpha/2} \) is the Z-score corresponding to the desired confidence level (99% in this case),
- \( \sigma \) is the standard deviation (21 minutes),
- \( E \) is the margin of error (± 5 minutes).

### Step-by-step solution:

1. **Z-score for 99% confidence**: 
   For 99% confidence, \( \alpha = 0.01 \), so \( \alpha/2 = 0.005 \). The Z-score corresponding to 0.005 in the standard normal distribution is approximately **2.576**.

2. **Substitute the values into the formula**:
   \[
   n = \left( \frac{2.576 \cdot 21}{5} \right)^2
   \]

3. **Calculate the required sample size**:

\[
n = \left( \frac{54.096}{5} \right)^2 = (10.8192)^2 = 117.024
\]

4. **Round up** to the nearest integer:  
   Since the sample size must be a whole number, we round up to **118**.

### Answer:
The required sample size is **118**.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How does increasing the confidence level affect the sample size?
2. What happens to the required sample size if the margin of error decreases?
3. Why do we round up the sample size instead of down?
4. How would a change in standard deviation impact the required sample size?
5. What are the assumptions behind using this formula for sample size calculation?

**Tip**: When calculating sample sizes, always round up to ensure the confidence level and margin of error are maintained!

An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous studies, the standard deviation is assumed to be 21 minutes. The executive wants to estimate, with 99% confidence, the mean weekly amount of time to within ± 5 minutes.

What sample size is needed?

Determine the required sample size to estimate the mean weekly time consumers spend watching traditional television with 99% confidence. Using a standard deviation of 21 minutes and a margin of error of ± 5 minutes, this statistical problem demonstrates key concepts like confidence intervals and Z-distribution.

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