Mathematics, 18.12.2019 18:31 felalv4444

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).

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Answer from: jaylene125

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 from: Quest

Volume = the base times the height. i’m not going your work for you, but this is all you need to know.

Answer from: Quest

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

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