Question

noted psychic was tested for extrasensory perception (ESP). The psychic was presented with 400 cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, or square. The psychic was correct in 120 cases. Let p represent the probability that the psychic correctly identified the symbols on the cards in a random trial. How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence

Answer 1

Answer:

A sample of 11352 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

400 cards, 120 cases were correct. So

$p = \frac{120}{400} = 0.3$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

How large a sample n would you need to estimate p with margin of error 0.01 and 95% confidence

We would need a sample of size n, and n is found when $M = 0.01$. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 2.325\sqrt{\frac{0.3*0.7}{n}}$

$0.01\sqrt{n} = 2.325\sqrt{0.3*0.7}$

$\sqrt{n} = \frac{2.325\sqrt{0.3*0.7}}{0.01}$

$(\sqrt{n})^2 = (\frac{2.325\sqrt{0.3*0.7}}{0.01})^2$

$n = 11351.8$

Rounding up

A sample of 11352 is needed.