3. Suppose a researcher is interested in the random variable detailing average d

Question

3. Suppose a researcher is interested in the random variable detailing average distance for patients traveled to the nearest health center decides to carry out a Z test. He collects a simple random sample of 100 patients sampled from a population with a known standard deviation of 32.8 miles. A) If the researcher wants to be 95% confident about his estimate, calculate the margin of error based on the above information. B) Carry out the same calculation above, but assuming the researcher wishes to be "virtually certain" about his estimate. (Use Z-3.0) C) Compare the margin of errors calculated above and explain the diffèerence.

Explanation / Answer

Solution:

Margin of error = zalpha/2 *SD/sqrt(n)

Margin of error = 1.96*32.8/sqrt(100)

Margin of error =1.96*32.8/10

Margin of error = 1.96*3.28

= 6.4288

Margin of error at 95% confidence interval is 6.4288

Solution2:

At Z=3 margin of error is

=3*32.8/sqrt(100)

=3*3.28

=9.84

Solution3

Margin of error at 95% confidence interval is 6.4288

And margin of error at 99.67% or Z=3 is 9.84 which contains 99.67% values. Where margin of error at 95% or Z=1.96 is 6.4288 which contains 95% values in Mean+-Margin of error.

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