Assume that 19.5% of people have sleepwalked. Assume that in a random sample of 1512 adults,...
Assume that 19.5% of people have sleepwalked. Assume that in a random sample of 1512
adults, 338 have sleepwalked. a. Assuming that the rate of 19.5% is correct, find the probability that 338 or more of the
1512 adults have sleepwalked. b. Is that result of 338 or more significantly high?
c. What does the result suggest about the rate of 19.5%
19.519.5%?
Homework Answers
a)
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.2235 - 0.195)/sqrt(0.195*(1-0.195)/1512)
z = 2.8
P(z > 2.8) = 0.0026
b)
Yes the result is significantly high
c)
The rate of people sleepwalked is greater than 19.5%
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