Question

The union for a particular industry has determined that the standard deviation of the daily wages of its workers is $20. A random sample of 90 workers in this industry has a mean daily wage of $122. Find a 90% confidence interval for the true mean daily wage of all union workers in the industry. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

Answer 1

Answer:

The 90% confidence interval for the true mean daily wage of all union workers in the industry is ($118.6, $125.4). The lower limit is $118.6 and the upper limit is $125.4.

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:

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

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 $1 - 0.05 = 0.95$, 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.

$M = 1.645\frac{20}{\sqrt{90}} = 3.4$

The lower end of the interval is the sample mean subtracted by M. So it is 122 - 3.4 = $118.6

The upper end of the interval is the sample mean added to M. So it is 122 + 3.4 = $125.4

The 90% confidence interval for the true mean daily wage of all union workers in the industry is ($118.6, $125.4). The lower limit is $118.6 and the upper limit is $125.4.