Question

If n=40, S.D=2.06 then the maximum error with 99% confidence is​

Answer 1

Answer:

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Answer 2

The maximum error with 99% confidence is 0.8387

Explanation:

Given

No. of samples $n = 40$

Standard Deviation $\sigma=2.06$

We know that z* for 99% confidence is 2.575

The maximum error is given by

$E=z*\frac{\sigma}{\sqrt{n}}$

$\implies E=2.575\times\frac{2.06}{\sqrt{40}}$

$\implies E=2.575\times\frac{2.06}{6.3246}$

$\implies E=0.8387$

Hope this answer is helpful.

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