An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 25 hours. How large a sample is needed if we wish to be 94% confident that our sample mean will be within 12 hours of the true mean?

MATLAB: An Introduction with Applications
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**Problem Statement:**

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 25 hours. How large a sample is needed if we wish to be 94% confident that our sample mean will be within 12 hours of the true mean?

**Explanation:**

This problem involves determining the sample size required for a confidence interval around a sample mean. The firm wants to be 94% confident that the sample mean is within 12 hours of the true mean. Given:

- Standard deviation (σ) = 25 hours
- Confidence level = 94%
- Margin of error (E) = 12 hours

**Approach:**

To find the sample size (n), the formula for the sample size of a mean in a normally distributed population is used:

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

Where:
- \( Z \) is the Z-score corresponding to the confidence level.
- \( \sigma \) is the standard deviation.
- \( E \) is the margin of error.

To solve, you would:
1. Find the Z-score for a 94% confidence level from Z-tables or standard normal distribution tables.
2. Substitute the Z-score, standard deviation (25 hours), and margin of error (12 hours) into the formula to calculate the required sample size.

This calculation will ensure precision in predicting the mean life of the light bulbs produced by the firm.
Transcribed Image Text:**Problem Statement:** An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 25 hours. How large a sample is needed if we wish to be 94% confident that our sample mean will be within 12 hours of the true mean? **Explanation:** This problem involves determining the sample size required for a confidence interval around a sample mean. The firm wants to be 94% confident that the sample mean is within 12 hours of the true mean. Given: - Standard deviation (σ) = 25 hours - Confidence level = 94% - Margin of error (E) = 12 hours **Approach:** To find the sample size (n), the formula for the sample size of a mean in a normally distributed population is used: \[ n = \left( \frac{{Z \times \sigma}}{E} \right)^2 \] Where: - \( Z \) is the Z-score corresponding to the confidence level. - \( \sigma \) is the standard deviation. - \( E \) is the margin of error. To solve, you would: 1. Find the Z-score for a 94% confidence level from Z-tables or standard normal distribution tables. 2. Substitute the Z-score, standard deviation (25 hours), and margin of error (12 hours) into the formula to calculate the required sample size. This calculation will ensure precision in predicting the mean life of the light bulbs produced by the firm.
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