Question

34 packages are randomly selected from packages received by a parcel service. The sample has a mean weight of 25.9 pounds and a standard deviation of 3.8 pounds. What is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?
24.6 < μ < 27.2

24.8 < μ < 27.0

24.2 < μ < 27.6

24.4 < μ < 27.4

Answer 1

Answer:

24.6 < μ < 27.2

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 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.

$M = 1.96\frac{3.8}{\sqrt{34}} = 1.3$

The lower end of the interval is the sample mean subtracted by M. So it is 25.9 - 1.3 = 24.6 pounds.

The upper end of the interval is the sample mean added to M. So it is 25.9 + 1.3 = 27.2 pounds.

So the correct answer is:

24.6 < μ < 27.2