Among 420 randomly selected employees at a company, the mean number of hours of overtime worked per month is 10 hours and the st

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2 years ago

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andard deviation is 1. 6. What is the margin of error, assuming a 99% confidence level? 4. 12 0. 01 0. 20 20. 5.

2 answers:

kari74 [83] · Answer 1 · 2022-12-31T07:33:11+03:00

The margin of error of the random selection is 0.20

The given parameters are:

$n = 420$ --- the sample size

$\sigma = 1.6$ --- the standard deviation

$\bar x = 10$ --- the mean

$\alpha = 99\%$ --- the confidence level.

The margin of error (E) is calculated as follows:

$E = z \times \sqrt{\frac{\sigma^2}{n}}$

So, we have:

$E = z \times \sqrt{\frac{1.6^2}{420}}$

$E = z \times \sqrt{\frac{2.56}{420}}$

The z-value for 99% confidence level is 2.576.

Substitute 2.576 for z

$E = 2.576 \times \sqrt{\frac{2.56}{420}}$

$E = 2.576 \times \sqrt{0.006095}$

Take square roots

$E = 2.576 \times 0.0781$

Multiply

$E = 0.2012$

Approximate

$E = 0.20$

Hence, the margin of error is 0.20