Mathematics, 25.02.2020 03:31 keviongardner

The Pew Research Center has conducted extensive research on the young adult population (Pew Research website, November 6, 2012). One finding was that 93% of adults aged 18 to 29 use the Internet. Another finding was that 21% of those aged 18 to 28 are married. Assume the sample size associated with both findings is 500.

Round your answers to four decimal places.

a.) Develop a 95% confidence interval for the proportion of adults aged 18 to 29 that use the Internet.

b.)Develop a 99% confidence interval for the proportion of adults aged 18 to 28 that are married

c.) in which case part (a) or (b) is the margin of error larger?

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Answer from: Nbeans4962

a) $0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

b) $0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

c) The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

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:

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

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:

$0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

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:

$0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

Part c

The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

Answer from: Quest

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