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

Answer 2

The 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service is;

Option A; 24.6 < μ < 27.2

We are given;

Sample mean; x' = 25.9 pounds

Standard deviation; σ = 3.8 pounds

Confidence level; CL = 95%

Sample size; n = 34

Now formula for confidence interval is;

CI = x' ± z( σ/√n)

Where z is critical value at given confidence level.

z at CL of 95% is; z = 1.96

Thus;

CI = 25.9 ± 1.96(3.8/√34)

CI = 25.9 ± 1.3

CI = (25.9 - 1.3), (25.9 + 1.3)

CI = (24.6, 27.2)

Read more at; https://brainly.com/question/12970960