Consider a population having a standard deviation equal to 10.07. We wish to est

Question

Consider a population having a standard deviation equal to 10.07. We wish to estimate the mean of this population. (a) How large a random sample is needed to construct a 95 percent confidence interval for the mean of this population with a margin of error equal to 1? (Round your answer to the next whole number.) (b) Suppose that we now take a random sample of the size we have determined in part a. If we obtain a sample mean equal to 277, calculate the 95 percent confidence interval for the population mean. What is the interval's margin of error? (Round your answers to the nearest whole number.)

Explanation / Answer

Z = 1.96 for 95% CI

Margin of error = 1

Standard deviation = 10.07

a.

The random sample is ?

1= 1.96*10.07/sqrt(n)

n = (10.07*1.96/1)^2

n~= 390

b.

The 95% CI will be :

277 - 1.96*10.07/sqrt(390) to 277 + 1.96*10.07/sqrt(390)

= 276 to 278

Margin of error = Z*sigma/sqrt(n) = 1.96*10.07/sqrt(390) = 1

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