Question

A measurement of fluoride ion in tooth paste from 5 replicate measurements delivers a mean of 0.14 % and a standard deviation of 0.05 %. What is the confidence interval at 95 % for which we assume that it contains the true value?

Answer 1

Answer:

The confidence interval at 95 % for which we assume that it contains the true value is (0.0962, 0.1838).

Step-by-step explanation:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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 M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

$M = 1.96*\frac{0.05}{\sqrt{5}} = 0.0438$

The lower end of the interval is the mean subtracted by M. So it is 0.14 - 0.0438 = 0.0962

The upper end of the interval is the mean added to M. So it is 0.14 + 0.0438 = 0.1838

The confidence interval at 95 % for which we assume that it contains the true value is (0.0962, 0.1838).