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shutvik [7]
2 years ago
6

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

andard deviation is 1. 6. What is the margin of error, assuming a 99% confidence level? 4. 12 0. 01 0. 20 20. 5.
Mathematics
2 answers:
kari74 [83]2 years ago
8 0

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

Read more about margin of error at:

brainly.com/question/14396648

Fittoniya [83]2 years ago
7 0

Answer:

0.20

Step-by-step explanation:

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Confidence interval:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

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