Consider a population having a standard deviation equal to 9.87. We wish to esti
ID: 3380265 • Letter: C
Question
Consider a population having a standard deviation equal to 9.87. 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 327, 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.)
Margin of error _______________
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)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
s = sample standard deviation = 9.87
E = margin of error = 1
Thus,
n = 374.2230098
Rounding up,
n = 375 [ANSWER]
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b)
Note that
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 327
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 9.87
n = sample size = 375
Thus,
Lower bound = 326.0010365
Upper bound = 327.9989635
Thus, the confidence interval is
( 326.0010365 , 327.9989635 ) [ANSWER]
Margin or error = (upper-lower)/2 = 0.998963476 [ANSWER]
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