Sleepwalking Assume thatof people have sleepwalked (based on "Prevalence and Comorbidity of Nocturnal Wandering in the U.S. Adult General Population," by Ohayon et al., Neurology, Vol. 78, No. 20).Assume that in a random sample of 1480 adults, 455 have sleepwalked.

a. Assuming that the rate ofis correct, find the probability that 455 or more of the 1480 adults have sleepwalked.

b. Is that result of 455 or more significantly high?

c. What does the result suggest about the rate of?

It is given that 29.2% of people sleepwalk. A sample of 1480 adults is considered, out of which 455 have sleepwalked.

Let X (number of successes) denote the number of people who have sleepwalked.

The probability of success is given to be equal to:

$$\begin{gathered} p=29.2\% \\ =\frac{29.2}{100} \\ =0.292 \end{gathered}$$

The number of trials (n) is equal to 1480.

Here, the sample is a result of 1480 independent trials with a probability of success at each trial equal to 0.292.

Also,

$$\begin{gathered} np=1480\times 0.292 \\ =432.16 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=n(1-p) \\ =1480\times (1-0.292) \\ =1047.84 \\ >5 \end{gathered}$$

Since the above two requirements are met, the normal distribution can be used for approximating the binomial distribution.

It is required to compute the probability of 455 or more adults who sleepwalk.

Thus, the interval of continuity correction is computed below:

$$\begin{gathered} \left(x-0.5,x+0.5\right)=\left(455-0.5,455+0.5\right) \\ =\left(454.5,455.5\right) \end{gathered}$$

In terms of the bound of the continuity correction interval, the following probability needs to be computed:

$P\left(x>\text{lower}\text{bound}\right)=P\left(x>454.5\right)$

The sample value equal to x=455 is converted to a z-score as follows:

$$\begin{gathered} z=\frac{x-np}{\sqrt{np(1-p)}} \\ =\frac{454.5-1480(0.292)}{\sqrt{1480(0.292)(1-0.292)}} \\ =1.27716 \\ \approx 1.28 \end{gathered}$$

a.

The probability of 455 or more adults who have sleepwalked is computed using the standard normal table as:

$$\begin{gathered} P\left(x>454.5\right)=P\left(z>1.28\right) \\ =1-P\left(z<1.28\right) \\ =1-0.8997 \\ =0.1003 \end{gathered}$$

Thus, the probability of 455 or more adults who have sleepwalked is equal to 0.1003.

b.

It can be said that the result of 455 or more people who have sleepwalked is not significantly high as the probability value is not small (less than 0.05).

c.

This suggests that there is not enough evidence to suggest that the true rate of people who sleepwalk is not equal to 29.2%.