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 symbol 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.Suppose you wished to see if there was evidence that the psychic is doing better than just guessing.To do this,you test the hypotheses H₀: p = 0.20 versus H_a: p > 0.20.What is the value of the large-sample z statistic?

A) z = 0.83
B) z = 1.77
C) z = 1.96
D) z = 4.17

Question

Janelle Jankowski · Accepted Answer

Final Answer : B, Explanation :   The large-sample z statistic is calculated using the formula   z = p ^ − p 0 p 0 ( 1 − p 0 ) n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} z = n p 0 ​ ( 1 − p 0 ​ ) ​ ​ p ^ ​ − p 0 ​ ​  , where   p ^ \hat{p} p ^ ​   is the sample proportion,   p 0 p_0 p 0 ​   is the null hypothesis proportion, and   n n n   is the sample size. Here,   p ^ = 50 200 = 0.25 \hat{p} = \frac{50}{200} = 0.25 p ^ ​ = 200 50 ​ = 0.25  ,   p 0 = 0.20 p_0 = 0.20 p 0 ​ = 0.20  , and   n = 200 n = 200 n = 200  . Plugging these values into the formula gives   z = 0.25 − 0.20 0.20 ( 0.80 ) 200 = 0.05 0.16 200 = 0.05 0.0008 = 0.05 0.0283 ≈ 1.77 z = \frac{0.25 - 0.20}{\sqrt{\frac{0.20(0.80)}{200}}} = \frac{0.05}{\sqrt{\frac{0.16}{200}}} = \frac{0.05}{\sqrt{0.0008}} = \frac{0.05}{0.0283} \approx 1.77 z = 200 0.20 ( 0.80 ) ​ ​ 0.25 − 0.20 ​ = 200 0.16 ​ ​ 0.05 ​ = 0.0008 ​ 0.05 ​ = 0.0283 0.05 ​ ≈ 1.77  .

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