Question

We would like to estimate the proportion of UF students who owns a scooter to within 1% of the truth, with 95% confidence. We believe around 10% of students do. How many students should be sampled? Group of answer choices 3458 34575 38416 3842

Answer 1

Answer:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.1(1-0.1)}{(\frac{0.01}{1.96})^2}=3457.44$  

And rounded up we have that n=3458

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (a)  

And on this case we have that $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.1(1-0.1)}{(\frac{0.01}{1.96})^2}=3457.44$  

And rounded up we have that n=3458

Answer 2

Using the confidence interval for a proportion, it is found that 3458 students should be sampled.

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this problem:

• 95% confidence, thus $\alpha = 0.95$, z has a p-value of $\frac{1 + 0.95}{2} = 0.975$, thus z = 1.96.
• Estimative of 10%, thus $p = 0.1$.

• The sample size is n, considering $M = 0.01$.

Then:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.1(0.9)}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.1(0.9)}$

$\sqrt{n} = \frac{1.96\sqrt{0.1(0.9)}}{0.01}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.1(0.9)}}{0.01}\right)^2$

$n = 3457.4$

Rounding up:

3458 students should be sampled.

A similar problem is given at https://brainly.com/question/13691929