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A researcher is interested in estimating the mean blood alcohol content of people arrested for driving under the influence. Based on past data, the researcher assumes a population standard deviation of 0.065. What sample size is needed to estimate the true mean blood alcohol content within .005 units at the 95% confidence level?

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Answer:

We need a sample size of at least 650

Explanation:

We have that to find our
\alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:


\alpha = (1-0.95)/(2) = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of
1-\alpha.

So it is z with a pvalue of
1-0.025 = 0.975, so
z = 1.96

Now, find the margin of error M as such


M = z*(\sigma)/(√(n))

In which
\sigma is the standard deviation of the population and n is the size of the sample.

In this problem:

We need a sample of size at least n

n is found when
M = 0.005, \sigma = 0.065

So


M = z*(\sigma)/(√(n))


0.005 = 1.96*(0.065)/(√(n))


0.005√(n) = 1.96*0.065


√(n) = (1.96*0.065)/(0.005)


(√(n))^(2) = ((1.96*0.065)/(0.005))^(2)


n = 649.23

Rounding up

We need a sample size of at least 650

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