Question

The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).

Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).

Answer 1

Answer:

a) The 95% confidence interval would be given by (0.4239;0.5161)

b) The 99% confidence interval would be given by (0.4093;0.5307)

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

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

$\hat p =0.47$ represent the estimated proportion of interest

n=450 represent the sample size

1) Part a

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 confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

$0.47 - 1.96\sqrt{\frac{0.47(1-0.47)}{450}}=0.4239$

$0.47 + 1.96\sqrt{\frac{0.47(1-0.47)}{450}}=0.5161$

The 95% confidence interval would be given by (0.4239;0.5161)

2) Part b

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

$0.47 - 2.58\sqrt{\frac{0.47(1-0.47)}{450}}=0.4093$

$0.47 + 2.58\sqrt{\frac{0.47(1-0.47)}{450}}=0.5307$

The 99% confidence interval would be given by (0.4093;0.5307)

Answer 2

Using the z-distribution, as we are working with a proportion, it is found that:

• The 95% confidence interval is (0.4239, 0.5161).

• The 99% confidence interval is (0.4094, 0.5306).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The parameters are given by: $\pi = 0.47, n = 450$.

For the 95% confidence interval,$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

Hence:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.4239$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 1.96\sqrt{\frac{0.47(0.53)}{450}} = 0.5161$

For the 99% confidence interval,$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so $z = 2.527$.

Hence:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 - 2.575\sqrt{\frac{0.47(0.53)}{450}} = 0.4094$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.47 + 2.575\sqrt{\frac{0.47(0.53)}{450}} = 0.5306$

To learn more about the z-distribution, you can check https://brainly.com/question/25890103