Question

Assume that 28.5% of people have sleepwalked. Assume that in a random sample of 1503 adults, 470 have sleepwalked. a. Assuming that the rate of 28.5% is correct, find the probability that 470 or more of the 1503 adults have sleepwalked. b. Is that result of 470 or more significantly high? c. What does the result suggest about the rate of 28.5%? . a. Assuming that the rate of 28.5% is correct, the probability that 470 or more of the 1503 adults have sleepwalked is (Round to four decimal places as needed.)

Answer 1

Answer

Given that

Assume that 28.5% of people have sleepwalked. Assume that in a random sample of 1503 adults, 470 have sleepwalked.

(a) . Assuming that the rate of 28.5% is correct, find the probability that 470 or more of the 1503 adults have sleepwalked.

Given that p = 28.5/100 = 0.285

sample size n = 1503

using normal approximation to binomial, we get

mean = n*p = 1503*0.285 = 428.355

and standard deviation

= Vn*p* (1-P)

= 1503 * 0.285 * (1 -0.285)

= V1503 +0.285 * 0.715

= 17.501

By continuity correcttion, we can write P(X\ge 470) = P(X>470-0.5) =P(X>469.5)

Using the formula P(X>x) = P(z>(x-mean)/sd)

= P(X>469.5)

=P(z>(469.5- 428.355)/17.501)

=P(z>2.35)

= 1 -P(z<2.35)

= 1 - 0.9906........(using z table)

= 0.0094

(b)  Is that result of 470 or more significantly high?

Yes, it is significantly high because the probability of 470 or more is less than 0.05 level.

(c) What does the result suggest about the rate of 28.5%?

Data suggests that the rate of 28.5% is lower because the actual percentage of adults who have sleepwalked is higher than 28.5%