Question

You measure the weights of a random sample of 400 male workers in the automotive industry. The sample mean is = 176.2 lbs. Suppose that the weights of male workers in the automotive industry follow a normal distribution with unknown mean μ and standard deviation σ = 11.1 lbs. A 95% confidence interval for μ is:

Answer 1

Answer:

A 95% confidence interval for μ is (175.287 lbs, 177.113 lbs).

Step-by-step explanation:

By the Central Limit Theorem, the mean of the sample is the same as the mean of the population. So:

Building the confidence interval:

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 $1-\alpha$.

So it is z with a pvalue of $1-0.05 = 0.95$, so $z = 1.645$

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 length of the sample. So

$M = 1.645*\frac{11.1}{\sqrt{400}}= 0.9130$

The lower end of the interval is the mean subtracted by M. So it is 176.2 - 0.9130 = 175.287 lbs

The upper end of the interval is the mean added to M. So it is 176.2 + 0.9130 = 177.113 lbs.

A 95% confidence interval for μ is (175.287 lbs, 177.113 lbs).