Question

1) A noted psychic was tested for ESP. The psychic was presented with 180 cards face down and was asked to determine if the card was one of 5 symbols: a star, cross, circle, square, or three wavy lines. The psychic was correct in 43 cases. Let p represent the probability that the psychic correctly identifies the symbol on the card in a random trial. Assume the 180 trials can be treated as an SRS from the population of all guesses.
To see if there is evidence that the psychic is doing better than just guessing, we test

H0:p=.2
??:?>.2

(a) What is the z-statistic for this test?  

(b) What is the P-value of the test?  

2) An Office of Admission document at a certain university claims that 56.1% of their undergraduates are female. To test this claim, a random sample of 230 undergraduates was selected. In this sample, 58.1% were female. Is there sufficient evidence to conclude that the document's claim is false? Carry out a hypothesis test at a 10% significance level.

A. The value of the test statistic is:

B. The p-value is

D. Your decision for the hypothesis test:

A. Reject H0.
B. Do Not Reject H0.
C. Reject H1.
D. Do Not Reject H1.

Answer 1

1)

Given :

(a) z-test statistic formula is

Therefore the z-statistic for this test is 1.30

(b)

p-value :

As the alternative hypothesis (Ha) is greater than (>) type , this is right tailed test .

So p-value = P(Z>z)

= 1 - P(Z<z)

Using Excel function   , =NORMSDIST (z)

=NORMSDIST(1.30)

=.9032

P-value = 1 - .9032 = .0968

P-value = .0968

Therefore the P-value of the test is .0968