An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of...

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An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of...

An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of n=82 n=82 C of I students found 30 with brown eyes. We test H 0 :p=.45 H0:p=.45 H a :p≠.45 Ha:p≠.45

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

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

please show what functions and steps you take , thank you

math Statistics-And-Probability

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Answer 1

Answer #1

a)

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.3659 - 0.45)/sqrt(0.45*(1-0.45)/82)
z = -1.53

b)

P-value Approach
P-value = 0.1260

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Answer 2

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