Mathematics, 27.02.2020 22:57 dee1334

In 2004, ACT, Inc. reported that 74% of 1644 randomly selected college freshmen returned to college the next year. The study was stratified by type of college-public or private. The retention rates were 71.9% among 505 students enrolled in public colleges and 74.9% among 1139 students enrolled in private colleges. a) Will the 95% confidence interval for the true national retention rate in private colleges be wider or narrower than the 95% confidence interval for the retention rate in public colleges? Explain. b) Do you expect the margin of error for the overall retention rate to be larger or smaller? Explain

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Answer from: deadslinger5134

a) Private colleges

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:

$0.749 - 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.724$

$0.749 + 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.774$

The 95% confidence interval would be given by (0.724;0.774)

The width of the interval is: 0.774-0.724=0.05

Public colleges

$0.719 - 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.680$

$0.719 + 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.758$

The 95% confidence interval would be given by (0.680;0.758)

The width of the interval is: 0.758-0.680=0.078

So then we can conclude that for public colleges we have a wider interval

b) Private colleges

$ME= 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}= 0.025$

Public colleges

$ME= 1.96\sqrt{\frac{0.719(1-0.719)}{505}}= 0.039$

We see a larger margin of error for the public colleges

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

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

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:

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

Private colleges

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:

$0.749 - 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.724$

$0.749 + 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}=0.774$

The 95% confidence interval would be given by (0.724;0.774)

The width of the interval is: 0.774-0.724=0.05

Public colleges

$0.719 - 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.680$

$0.719 + 1.96\sqrt{\frac{0.719(1-0.719)}{505}}=0.758$

The 95% confidence interval would be given by (0.680;0.758)

The width of the interval is: 0.758-0.680=0.078

So then we can conclude that for public colleges we have a wider interval

Part b

Private colleges

$ME= 1.96\sqrt{\frac{0.749(1-0.749)}{1139}}= 0.025$

Public colleges

$ME= 1.96\sqrt{\frac{0.719(1-0.719)}{505}}= 0.039$

We see a larger margin of error for the public colleges

Answer from: Quest

step-by-step explanation:

Answer from: Quest

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In 2004, ACT, Inc. reported that 74% of 1644 randomly selected college freshmen returned to college...