The margin of error of the random selection is 0.20
The given parameters are:
[tex]n = 420[/tex] --- the sample size
[tex]\sigma = 1.6[/tex] --- the standard deviation
[tex]\bar x = 10[/tex] --- the mean
[tex]\alpha = 99\%[/tex] --- the confidence level.
The margin of error (E) is calculated as follows:
[tex]E = z \times \sqrt{\frac{\sigma^2}{n}}[/tex]
So, we have:
[tex]E = z \times \sqrt{\frac{1.6^2}{420}}[/tex]
[tex]E = z \times \sqrt{\frac{2.56}{420}}[/tex]
The z-value for 99% confidence level is 2.576.
Substitute 2.576 for z
[tex]E = 2.576 \times \sqrt{\frac{2.56}{420}}[/tex]
[tex]E = 2.576 \times \sqrt{0.006095}[/tex]
Take square roots
[tex]E = 2.576 \times 0.0781[/tex]
Multiply
[tex]E = 0.2012[/tex]
Approximate
[tex]E = 0.20[/tex]
Hence, the margin of error is 0.20
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