Find the sample size required to estimate the percentage of college students who use student loans to help fund their tuition. Assume that we want 95% confidence that the proportion from the sample is within two percentage points of the true population percentage. Record your answer as a whole number.

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

Assume that we want 95% confidence that the proportion from the sample is within two percentage points of the true population percentage.

We need a sample of size at least n.

n is found when [tex]M = 0.02[/tex]

We don't know the exact proportion, so we use [tex]\pi = 0.5[/tex], which is the case for which we are going to need the largest sample size.

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.02\sqrt{n} = 1.96*0.5[/tex]

[tex]\sqrt{n} = \frac{1.96*0.5}{0.02}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}[/tex]

[tex]n = 2401[/tex]

We need a sample of size at least 2401.

dfbustos dfbustos · Answer 2 · 2020-04-23T02:41:29+02:00

Answer:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401[/tex]

And rounded up we have that n=2401

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".

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 [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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

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

And on this case we have that [tex]ME =\pm 0.02[/tex] or 2% points and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

We can use an prior estimation for p [tex]\hat p=0.5[/tex]. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401[/tex]

And rounded up we have that n=2401