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 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.

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?

Find out how to calculate the sample size needed to estimate the mean weekly amount of time spent watching TV with 99% confidence, given a standard deviation of 21 minutes and a margin of error of ±5 minutes. This step-by-step guide covers statistical concepts and key formulas for sample size estimation.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation