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Xelga [282]
2 years ago
13

The recommended daily intake of vitamin C for males 18 years and older is 90 milligrams. If an 18 year-old male has already cons

umed 36 milligrams of vitamin C, what percentage of the daily recommended amount has he already consumed?
Mathematics
1 answer:
Ksju [112]2 years ago
6 0

Answer:

40%

Step-by-step explanation:

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Marizza181 [45]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

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

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 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

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 0.04 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)  

We assume that a prior estimation for p would be \hat p =0.5 since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

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Step-by-step explanation:


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