Mathematics: You have been assigned to determine whether more people prefer Coke or Pepsi. Assume that roughly half the population prefers Coke and half prefers Pepsi. How large a sample do you need to take to ensure that you can estimate, with 95% confidence, the proportion of people preferring Coke within 3% of the actual value? [Hint: proportion est. = 0.5] Round your answer to whole number

20:15 is the answer. A ratio is comparing one number the the other number, using :...

You have been assigned to determine whether more

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SMURFETTE86

29.01.2022

Mathematics

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

For this case we have the following info given:

ME=0.03 the margin of error desired

Conf= 0.95 the level of confidence given

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)

the critical value for 95% of confidence is z=1.96

We can use as estimator for the population of interest $\hat p=0.5$ . And on this case we have that $ME =\pm 0.03$ 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:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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