Mathematics, 24.12.2019 20:31 rpegler4

An investigator wants to estimate caffeine consumption in high school students.
a)- how many students would be required to ensure that a 95% confidence interval estimate for the mean caffeine intake (measured in mg) is within 15 units of the true mean? assume that the standard deviation in caffeine intake is 68mg

b)-how many students would be required to estimate the proportion of students who consume coffee? suppose we want the estimate to be within 5% of the true proportion with 95% confidence. show work.

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Answer from: brennarfa

a) $n=(\frac{1.960(68)}{15})^2 =78.94 \approx 79$

So the answer for this case would be n=79 rounded up to the nearest integer

b) $n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n represent the sample size

Part a

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =15 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(68)}{15})^2 =78.94 \approx 79$

So the answer for this case would be n=79 rounded up to the nearest integer

Part b

The population proportion have the following distribution

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

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 = 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*)

Since we don't have any info provided we can assume $\hat p =0.5$ . And replacing into equation (b*) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

Answer from: Quest

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

$(zy )^{2} +10^{2} = {(10 \sqrt{2}) }^{2}$

$(zy )^{2} +100= 200$

group similar terms to get,

$(zy )^{2}= 200 - 100$

this implies that,

$(zy )^{2}=100$

take positive square root of both sides to get,

$zy = \sqrt{100}$

this implies that,

$zy = 10$

What is the distance from point y to wx in the figure below

Answer from: Quest

did you get the answer yet?

step-by-step explanation:

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