A researcher wants to find the mean diameter of washers produced by a company. The sample size is 120 washers, with an average diameter of 28 inches. If the standard deviation is 0.5 and the confidence level is 99%, what is the margin of error of the mean? Round your answer to three decimal places.



Answer :

To find the margin of error for the mean diameter of the washers produced in the company, we will use the following steps:

1. Identify the given data:
- Sample size ([tex]\( n \)[/tex]): 120
- Sample mean ([tex]\( \bar{x} \)[/tex]): 28 inches
- Population standard deviation ([tex]\( \sigma \)[/tex]): 0.5 inches
- Confidence level: 99%

2. Determine the Z-score for the given confidence level:
For a 99% confidence level, we need to find the Z-score that corresponds to the middle 99% of the standard normal distribution. The remaining 1% is divided equally into the two tails of the distribution, so each tail has 0.5%.

To find the Z-score, we look up the value that corresponds to [tex]\( 1 - \frac{1-0.99}{2} = 0.995 \)[/tex] in the Z-table. This value is approximately 2.576.

3. Calculate the margin of error (E):
The formula for the margin of error is:
[tex]\[ E = Z \times \left( \frac{\sigma}{\sqrt{n}} \right) \][/tex]
Where:
- [tex]\( Z \)[/tex] is the Z-score
- [tex]\( \sigma \)[/tex] is the population standard deviation
- [tex]\( n \)[/tex] is the sample size

Plugging in the values:
[tex]\[ E = 2.576 \times \left( \frac{0.5}{\sqrt{120}} \right) \][/tex]

4. Simplify the calculation:
First, compute [tex]\( \frac{0.5}{\sqrt{120}} \)[/tex]:
[tex]\[ \frac{0.5}{\sqrt{120}} \approx \frac{0.5}{10.954} \approx 0.0457 \][/tex]

Then, multiply by the Z-score:
[tex]\[ E = 2.576 \times 0.0457 \approx 0.118 \][/tex]

5. Conclusion:
The margin of error of the mean diameter of the washers, at a 99% confidence level, is approximately 0.118 inches (rounded to three decimal places).

So, the margin of error of the mean is [tex]\(\mathbf{0.118}\)[/tex] inches.