To determine the required sample size for estimating the mean weekly amount of time spent watching traditional television with 99% confidence and within ± 5 minutes, we use the following formula for sample size estimation when the population standard deviation $\sigma$ is known:

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

Where:

• $Z_{\alpha/2}$ is the z-score corresponding to the 99% confidence level,
• $\sigma = 21$ minutes is the population standard deviation,
• $E = 5$ minutes is the margin of error.

Step 1: Find the z-score

For a 99% confidence level, $\alpha = 1 - 0.99 = 0.01$, so $\alpha/2 = 0.005$. The z-score for 99% confidence is approximately $Z_{\alpha/2} = 2.576$.

Step 2: Plug in the values

Now substitute the known values into the formula:

$n = \left( \frac{2.576 \cdot 21}{5} \right)^2$

Let's calculate this.

The required sample size for 99% confidence is 117.

Would you like further details or have any questions?

Here are some related questions to explore:

1. How does the confidence level affect the sample size?
2. What happens to the sample size if the margin of error is reduced to 3 minutes?
3. How does changing the standard deviation impact the required sample size?
4. What is the significance of the z-score in confidence interval estimation?
5. How can you estimate the sample size if the standard deviation is unknown?

Tip: A higher confidence level results in a larger required sample size for the same margin of error.