Question

One of the more famous anecdotes in the field of statistics is known as the "Lady Tasting Tea" and involves the renowned statistician Sir Ronald Aylmer Fisher. A woman claimed to be able to tell the difference in a cup of tea depending on whether or not the milk or tea was poured first. You have chosen to recreate this experience with a classmate and have fixed 40 cups of tea, of which your classmate correctly identifies 12. How large a sample n would you need to estimate p with margin of error 0.01 with 95% confidence? Use the guess p = 0.30 as the value for p. n = 42 n = 81 n = 4116 n = 8068

Answer 1

Answer:

$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$  

And rounded up we have that n=8068

Step-by-step explanation:

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

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

We want for this case a 95% of confidence desired, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

The estimated proportion for this case is $\hat p=0.3$. And replacing into equation (b) the values from part a we got:

$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$  

And rounded up we have that n=8068