Question

The blood alcohol content (BAC) of a person will rise after drinking alcoholic drinks such as beer or wine. Each person’s BAC responds differently based on many physiological factors. In Australia, a person must have a BAC lower than 0.05 in order to legally drive.

Your colleague suggests that you should compare the difference in BAC directly rather than whether the study subjects can legally drive or not. Suppose that the BAC of people under the study conditions who do not take the drug has an N(0.04, 0.012) distribution.

a. Suppose the drug reduces the mean BAC by α and does not change the variance. What value of α leads to a 0.1 increase in the proportion of people who can legally drive?

b. Assuming this is the true value of α, what value of n do you need in order for a 95% confidence interval to exclude zero? (Assume you know the common σ = 0.01, but not the population means.)

Answer 1

First we need to find P( X < 0.05)

Let's used excel:

P( X < 0.05) = "=NORMDIST(0.05,0.04,0.012,1)" = 0.7977.

Now we want to find such that P( X - < 0.05  ) = P( X < 0.05 + ​ ) = 0.7977 + 0.1 = 0.8977

Let's use excel:

0.05 + = "=NORMINV(0.8977,0.04,0.012)" = 0.055223

Therefore   = 0.055223 - 0.05 = 0.005223

b)

Formula of sample size calculation is as follow

For 95% confidence level Z_c = 1.96

E = margin of error = = 0.005223

= 0.01

Plug these values in the above formula of n, we get

So final answer is n = 15

(Note that we round n to the next whole number)