Mathematics, 21.04.2020 03:53 dewaynesmith46

A researcher wishes to estimate the proportion of adults who have high-speed internet access. what size sample should be obtained if she wishes the estimate to be within 0.050.05 with 9999% confidence if (a) she uses a previous estimate of 0.580.58? (b) she does not use any prior estimates?

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Answer from: Megcuttie101

a) $n=\frac{0.58(1-0.58)}{(\frac{0.05}{2.58})^2}=648.600$

And rounded up we have that n=649

b) $n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.58})^2}=665.64$

And rounded up we have that n=666

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 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:

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

Part a

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

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

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.58(1-0.58)}{(\frac{0.05}{2.58})^2}=648.600$

And rounded up we have that n=649

Part b

For this case we can use an estimator for p as $\hat p=0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.58})^2}=665.64$

And rounded up we have that n=666

Answer from: Quest

answer: 590

step-by-step explanation:

Answer from: Quest

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

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