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?

**b.** If​ 95% confidence is​ desired, how many consumers need to be​ selected?

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

enter your response here

​(Round up to the nearest​ integer.)

**b.** The sample size required for​ 95% confidence is

enter your response here

​(Round up to the nearest​ integer.)


To solve this problem, we need to use the formula for the sample size required for estimating the population mean with a specific level of confidence:

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

Where:
- \( Z_{\alpha/2} \) is the Z-score corresponding to the desired confidence level,
- \( \sigma \) is the population standard deviation,
- \( E \) is the margin of error.

### Given information:
- \( \sigma = 21 \) minutes,
- \( E = 5 \) minutes,
- Confidence levels: 99% and 95%.

#### a. Sample size for 99% confidence:

For 99% confidence, the Z-score \( Z_{\alpha/2} \) is approximately 2.576.

Substitute the values into the sample size formula:

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

#### b. Sample size for 95% confidence:

For 95% confidence, the Z-score \( Z_{\alpha/2} \) is approximately 1.96.

Substitute the values into the sample size formula:

\[
n = \left( \frac{1.96 \cdot 21}{5} \right)^2
\]

I'll calculate both sample sizes now.### a. The sample size required for 99% confidence is **118**.

### b. The sample size required for 95% confidence is **68**.

Would you like further details on the calculations or any clarifications?

Here are some related questions for further exploration:

1. How does changing the standard deviation affect the sample size required?
2. What happens to the sample size if the margin of error is increased to 10 minutes?
3. Why does the sample size differ for different confidence levels?
4. How is the Z-score determined for different confidence levels?
5. What impact does rounding the sample size have on the accuracy of the estimate?

**Tip**: When calculating sample sizes, always round up to ensure your confidence interval is accurate!

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?

**b.** If 95% confidence is desired, how many consumers need to be selected?

Learn how to calculate the sample size required to estimate the mean weekly TV viewing time of consumers with 99% and 95% confidence. The problem involves statistical concepts such as confidence intervals, Z-scores, and sample size formulas. Given a standard deviation of 21 minutes and a margin of error of ±5 minutes, this guide provides a step-by-step solution.

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