Question

g A psychic was tested for extrasensory perception (ESP). The psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, or square. 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:

The sample size should be 6157

Step-by-step explanation:

Given that the margin of error (e) = ± 0.01 and the confidence (C) = 95% = 0.95.

Let us assume that the guess p = 0.25 as the value of p.

α = 1 - C = 1 - 0.95 = 0.05

$\frac{\alpha }{2} =\frac{0.05}{2}=0.025$

The Z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is 1.96. Therefore $Z_\frac{\alpha }{2}=Z_{0.025}=1.96$

To determine the sample size n, we use the formula:

$Z_{0.025}*\sqrt{\frac{p(1-p)}{n} }\leq e\\Substituting:\\1.96*\sqrt{\frac{0.2(1-0.2)}{n} } \leq 0.01\\\sqrt{\frac{0.2(0.8)}{n} }\leq \frac{1}{196}\\\sqrt{0.16} *196 \leq \sqrt{n}\\78.4\leq \sqrt{n}\\ 6146.56\leq n\\n=6157$

Answer 2

Answer:

Step-by-step explanation:

Hello!

The objective is to test ESP, for this, a psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, square.

Be X: number of times the psychic identifies the symbols on the cards correctly is a size n sample.

p the probability that the psychic identified the symbol on the cards correctly

You have to calculate the sample size n to estimate the proportion with a confidence level of 95% and a margin of error of d=0.01

The CI for the population proportion is constructed "sample proportion" ± "margin of error" Symbolically:

p' ± $Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$

Where  $d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$ is the margin of error. As you can see, the formula contains the sample proportion (it is normally symbolized p-hat, in this explanation I'll continue to symbolize it p'), you have to do the following consideration:

Every time the psychic has to identify a card he can make two choices:

"Success" he identifies the card correctly

"Failure" he does not identify the card correctly

If we assume that each symbol has the same probability of being chosen at random P(star)=P(cross)=P(circle)=P(square)= 1/4= 0.25

Let's say, for example, that the card has the star symbol.

The probability of identifying it correctly will be P(success)= P(star)= 1/4= 0.25

And the probability of not identifying it correctly will be P(failure)= P(cross) + P(circle) + P(square)= 1/4 + 1/4 + 1/4= 3/4= 0.75

So for this experiment, we'll assume the "worst case scenario" and use p'= 1/4 as the estimated probability of the psychic identifying the symbol on the card correctly.

The value of Z will be $Z_{1-\alpha /2}= Z_{0.975}= 1.96$

Now using the formula you have to clear the sample size:

$d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$

$\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p')}{n} }$

$(\frac{d}{Z_{1-\alpha /2}})^2 =\frac{p'(1-p')}{n}$

$n*(\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')$

$n = p'(1-p')*(\frac{Z_{1-\alpha /2}}{d})^2$

$n = (0.25*0.75)*(\frac{1.96}{0.01})^2= 7203$

To estimate p with a margin of error of 0.01 and a 95% confidence level you have to take a sample of 7203 cards.

I hope this helps!