Question

A research study showed that a sample of 25 dosages from a medicine dispensing machine averaged 24.1 mg of a certain drug. The sample standard deviation was 1.8 mg. a) Form a 99% confidence interval of the population average of drug dosage dispensed. b) Explain what this interval means to someone unfamiliar with this research.

Answer 1

a) The 99% confidence interval of the population average of drug dosage dispensed is between 23.1 mg and 25.1 mg.

b) It means that we are 99% sure that the true mean dosage for all the population is between 23.1 mg and 25.1 mg.

Explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

a)

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 25 - 1 = 24

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 2.797

The margin of error is:

`M = T(s)/(√(n)) = 2.797(1.8)/(√(25)) = 1`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 24.1 - 1 = 23.1 mg

The upper end of the interval is the sample mean added to M. So it is 24.1 + 1 = 25.1 mg

The 99% confidence interval of the population average of drug dosage dispensed is between 23.1 mg and 25.1 mg.

b)

The interval means that we are 99% sure that the true mean dosage for all the population is between 23.1 mg and 25.1 mg.