Question

A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their purchasing habits. Construct the 80% confidence interval for the mean number of toys purchased each year if the sample mean was found to be 7.6. Assume that the population standard deviation is 1.5. Round your answers to one decimal place.

Answer 1

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

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:

$\alpha = \frac{1 - 0.8}{2} = 0.1$

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

That is z with a pvalue of $1 - 0.1 = 0.9$, so Z = 1.28.

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.

$M = 1.28\frac{1.5}{\sqrt{305}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.