Question

Can you taste PTC? 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 who have at least one Italian grandparent and who can taste PTC.

(a) How large a sample must you test to estimate the proportion of PTC tasters within $0.04$with $90$% confidence? Answer this question using the $75$% estimate as to the guessed value for $\stackrel{\wedge }{p}$.

(b) Answer the question in part (a) again, but this time use the conservative guess $\stackrel{\wedge }{p}=0.5$. By how much do the two sample sizes differ?

Short answer

From the given information,

a) The sample size of $318$has tested PTC when the sample proportion $\stackrel{\wedge }{p}is75\%$

b) The sample size is increased by$105$when the guessed value$\stackrel{\wedge }{p}is0.5$

Step-by-step solution

part (a) Step 1: Given Information 

It is given in the question that,

margin of error $E=0.04$

guessed value $\stackrel{\wedge }{p}=75\%$

confidence interval =$90\%$

part (a) Step 2: Explanation 

The confidence interval is $90\%$

convert $90\%$ into decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A

convert $75\%$into decimal.

$\frac{75}{100}=0.75$

Now, find the sample size. Use the formula $n=\frac{{\left[z_{a/2}\right]}^2\overrightarrow{p}(1-\overrightarrow{p})}{E^2}$

$n=\frac{{\left[z_{a/2}\right]}^2\overrightarrow{p}(1-\overrightarrow{p})}{E^2}$

$n=\frac{{1.645}^2\times 0.75\times (1-0.75)}{{0.04}^2}$

$=318$

Therefore, the sample size is$318$

part (b) Step 1: Given Information 

It is given in the question that,

margin of error $E=0.04$

guessed value $\stackrel{\wedge }{P}=75\%$

confidence interval =$90\%$

By how much do the two sample sizes differ?

part (b) Step 2: Explanation 

Convert $90\%$into decimal.

$\frac{90}{100}=0.90$

For confidence interval $0.90$, use table A.

$z_{\alpha /2}=1.645$

From part (a) sample size n = $318$

Find the new sample size. Use the formula $n=\frac{{\left[z_{o/2}\right]}^2\overrightarrow{p}(1-\overrightarrow{p})}{E^2}$

$n^{\mathrm{\prime }}=\frac{{\left[z_{a/2}\right]}^2\overrightarrow{p}(1-\overrightarrow{p})}{E^2}$

$n^{\mathrm{\prime }}=\frac{{1.645}^2\times 0.5\times (1-0.5)}{{0.04}^2}$

$=423$

Thus, the new sample size is $423$

For the difference between the sample sizes, subtract $n$from $n\text{'}$

$n^{\mathrm{\prime }}-n=423-318$

$=105$