Question

a noted psychic was tested for extrasensory perception. the psychic was presented with 200 cards face down and asked to determine if each card were one of five symbols: a star, a cross, a circle, a square, or three wavy lines. the psychic was correct in 50 cases.

let p represent the probability that the psychic correctly identifies the symbols on the card in a random trial. assume the 200 trials can be treated as a simple random sample from the population of all guesses the psychic would make in his lifetime.

1.based on the results what is a 95% confidence interval for p (use the large sample confidence interval)

a. 0.25 + or - 0.06

b. 0.25 plus or minus 0.004

c. 0.25 plus or minus 0.05

d. 0.25 + or - 0.055

2. suppose we now perform a hypothesis test with the no hypothesis being that the psychic has no ability so p equals 0.20. what is the p-value for the two-sided test?

a. 0.05 <P value<0.10

b. 0.01<P value <0.05

c. 0.001<P value<0.01

d. P value < 0.001

e. 0.10< P value

Answer 1

1)Here, the sample proportion , = 50 / 200

= 0.25

n = 200

z-value for 95% confidence interval = 1.96

Therefore, the confidence interval is = z*(1-)/n

= 0.25 1.96 * 0.0306

= 0.25 0.06

Answer) a. 0.25 + or - 0.06

2) to test,

H0 : p = 0.2

H1 : p 0.2

= 0..25

The test statistic , z = ( - p) / p(1-p)/n

= 0.05 / 0.028

= 1.786

We find the corresponding p-value for the z statistic using MS-Excel.

P-value = 0.074

Answer) a. 0.05 < p value < 0.10