PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inherited. About 75% of Italians can taste PTC, for example. You want to estimate the proportion of Americans with at least one Italian grandparent who can taste PTC. Starting with the 75% estimate for Italians, how large a sample must you test in order to estimate the proportion of PTC tasters within 0.032 with 80% confidence

And rounded up we have that n=300

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

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 80% of confidence, our significance level would be given by and . And the critical value would be given by:

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

   (a)  

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

  (b)  

The estimation for is given.

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

And rounded up we have that n=300

the answer is ! ?

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