A noted psychic was tested for ESP. The psychic was presented with 200 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 49 cases. Let pp represent the probability that the psychic correctly identifies the symbol on the card in a random trial. Assume the 200 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 testH0:p=.2H0:p=.2Ha:p>.2Ha:p>.2(a) What is the zz-statistic for this test? (b) What is the P-value of the test?

Answer:a. The Z-statistic is [tex]Z = \frac{\hat{p}-0.2}{\sqrt{0.2(0.8)/200}}[/tex] and the observed value 1.591 b. p-value: 0.0558Step-by-step explanation:To see if there is evidence that the psychic is doing better than just guessing, we want to test the next hypothesis [tex]H_{0}: p = 0.2[/tex] vs [tex]H_{a}: p > 0.2[/tex] (upper-tail alternative) We have a large sample size of n = 200 trials. Therefore, the test statistic is given by [tex]Z = \frac{\hat{p}-0.2}{\sqrt{0.2(0.8)/200}}[/tex] which is normally distributed. The observed value is  [tex]z = \frac{49/200-0.2}{\sqrt{0.2(0.8)/200}} = 1.591[/tex].  The p-value is calculated as P(Z > 1.591) = 0.0558. With this p-value we fail to reject the null hypothesis at the significance level of 0.05 for instance.