Consider a population having a standard deviation equal to 10.06. We wish to est

Question

Consider a population having a standard deviation equal to 10.06. We wish to estimate the mean of this population.

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.)

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 291, 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.)

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.)

Explanation / Answer

a.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 10.06
ME =1
n = ( 1.96*10.06/1) ^2
= (19.718/1 ) ^2
= 388.784 ~ 389      

b.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=291
Standard deviation( sd )=10.06
Sample Size(n)=389
Confidence Interval = [ 291 ± Z a/2 ( 10.06/ Sqrt ( 389) ) ]
= [ 291 - 1.96 * (0.51) , 291 + 1.96 * (0.51) ]
= [ 290,292 ]

Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value

Margin of Error = Z a/2 * 10.06/ Sqrt ( 389)
= 1.96 * (0.5101)
= 0.9997

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