Question

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 76% of people from a certain country can taste PTC, for example. You want to estimate the proportion of people with at least one grandparent from this country who can taste PTC. (Round your answers up to the nearest person.)

(a) Starting with the 76% estimate, how large a sample must you collect in order to estimate the proportion of PTC tasters within ±0.01 with 90% confidence?
(b) Estimate the sample size required if you made no assumptions about the value of the proportion who could taste PTC?

Answer 1

a)
`n=(0.76(1-0.76))/(((0.01)/(1.64))^2)=4905.83`

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion
`\hat p =0.5`

`n=(0.5(1-0.5))/(((0.01)/(1.64))^2)=6724`

And rounded up we have that n=6724

Explanation:

We need to remember that the confidence interval for the true proportion is given by :

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

Part a

The estimated proportion for this case is
`\hat p =0.76`

Our interval is at 90% of confidence, and the significance level is given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. The critical values for this case are:

`z_(\alpha/2)=-1.64, t_(1-\alpha/2)=1.64`

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

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

The margin of error desired is given
`ME =\pm 0.01` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

Replacing we got:

`n=(0.76(1-0.76))/(((0.01)/(1.64))^2)=4905.83`

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion
`\hat p =0.5`

`n=(0.5(1-0.5))/(((0.01)/(1.64))^2)=6724`

And rounded up we have that n=6724