Question

A psychiatrist is interested in finding a 90% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 13 randomly selected children with Tourette syndrome. The population standard deviation is σ=4.08.
Required:
a. To compute the confidence interval use a, z or t distribution?
b. With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between ______ and _______ .

Answer 1

Answer:

a) z-distribution is used, as we have the standard deviation for the population.

b) Between $\overline{x} - 1.86$ and $\overline{x} + 1.86$, in which $\overline{x}$ is the sample mean of tics per hour.

Step-by-step explanation:

a. To compute the confidence interval use a, z or t distribution?

We have the standard deviation for the population, and thus, the z-distribution is used.

b. With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between ______ and _______ .

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

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error 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.

$M = 1.645\frac{4.08}{\sqrt{13}} = 1.86$

The lower end of the interval is the sample mean subtracted by M. So it is $\overline{x} - 1.86$

The upper end of the interval is the sample mean added to M. So it is $\overline{x} + 1.86$

Between $\overline{x} - 1.86$ and $\overline{x} + 1.86$, in which $\overline{x}$ is the sample mean of tics per hour.