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' ± [tex]Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex]
Where [tex]d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex] 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 [tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]
Now using the formula you have to clear the sample size:
[tex]d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )[/tex]
[tex]\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p')}{n} }[/tex]
[tex](\frac{d}{Z_{1-\alpha /2}})^2 =\frac{p'(1-p')}{n}[/tex]
[tex]n*(\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')[/tex]
[tex]n = p'(1-p')*(\frac{Z_{1-\alpha /2}}{d})^2[/tex]
[tex]n = (0.25*0.75)*(\frac{1.96}{0.01})^2= 7203[/tex]
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!
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
[tex]\frac{\alpha }{2} =\frac{0.05}{2}=0.025[/tex]
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 [tex]Z_\frac{\alpha }{2}=Z_{0.025}=1.96[/tex]
To determine the sample size n, we use the formula:
[tex]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[/tex]
