At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.

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 $1-\alpha$ .

So it 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.

In this question:

We need a sample size of at least n.

n is found when M = 3. We have that $\sigma = 18$ .

$M = z*\frac{\sigma}{\sqrt{n}}$

$3 = 1.645*\frac{18}{\sqrt{n}}$

$3\sqrt{n} = 1.645*18$

Simplifying by 3

$\sqrt{n} = 1.645*6$

$(\sqrt{n})^{2} = (1.645*6)^{2}$

$n = 97.41$

Rounding up,

At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.

8 0

3 years ago

If Jose has a car with a mass of 55,000 kg and it is acted upon with a force of 200 Newtons. What is the acceleration?

zloy xaker [14]

Answer:

275 m/s2

Step-by-step explanation:

I know because I had this question in a test and got it correct.

6 0

3 years ago