SOLUTION: Assume that 32.9​% of people have sleepwalked. Assume that in a random sample of 1521 ​adults, 542 have sleepwalked. a. Assuming that the rate of 32.9​% is&#8203

Algebra.Com
Question 1125872: Assume that 32.9​% of people have sleepwalked. Assume that in a random sample of 1521 ​adults, 542 have sleepwalked.
a. Assuming that the rate of 32.9​% is​ correct, find the probability that 542 or more of the 1521 adults have sleepwalked.
b. Is that result of 542 or more significantly​ high?
c. What does the result suggest about the rate of 32.9​%?

Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
a. 549/1521=36.1%
z=(p-phat)/sqrt(p*(1-p)/n)
=0.032/sqrt(0.329*0.671/1521)
=2.66
b. probability z>2.66 is 0.0039
This is about 2/5 ths of one percent, considered to be statistically significant.
c. That suggests that perhaps the quoted rate of 32.9% is too low.

RELATED QUESTIONS

Assume that adults have IQ scores that are normally distributed with a mean of 105... (answered by Boreal)
Assume that adults have IQ scores that are normally distributed with a mean of 105... (answered by Boreal)
Assume that adults have IQ scores that are normally distributed with a mean of 100100 and (answered by Boreal)
In a random sample of 24 ​people, the mean commute time to work was 30.8 minutes... (answered by Boreal)
In a random sample of 23 ​people, the mean commute time to work was 34.4 minutes... (answered by Boreal)
A publisher wants to estimate the mean length of time​ (in minutes) all adults... (answered by Boreal)
A medical researcher says that less than 24​% of adults in a certain country are... (answered by Boreal)
In a random sample of four microwave​ ovens, the mean repair cost was... (answered by rothauserc)
Find the probability and interpret the results. If​ convenient, use technology to... (answered by oscargut)