Q. 7.50 - Elementary Statistics

Question

7.50 Undergraduate Binge Drinking. Alcohol consumption on college and university campuses has gained attention because undergraduate students drink significantly more than young adults who are not students. Researchers I. Balodis et al. studied binge drinking in undergraduates in the article "Binge Drinking in Undergraduates: Relationships with Gender, Drinking Behaviors, Impulsivity, and the Perceived Effects of Alcohol" (Behavioural Pharmacology, Vol. 20, No. 5. pp. 518-526). The researchers found that students who are binge drinkers drink many times a month with the span of each outing having a mean of  $4.9$ hours and a standard deviation of $1.1$ hours.
a. For samples of size $40$, find the mean and standard deviation of all possible sample mean spans of binge drinking episodes. Interpret your results in words.
b. Repeat part (a) with $n=120$.

Step-by-Step Solution

Answer

(a) The mean is $4.9$ hours and the standard deviation is $0.17$ hours.

(b) The mean is $4.9$ hours and the standard deviation is $0.1004$ hours.

Part (a) Step 1: Given information

To find the mean and standard deviation of all possible sample mean spans of binge drinking episodes for sample size $40$.

Part (a) Step 2: Explanation

The mean $\left(\mu \right)$span of binge drinking episodes is $4.9$ hours. The mean $\left({\mu }_X\right)$ for sample size $40$:
${\mu }_x=\mu$
${\mu }_x=4.9$
For sample size 40, the standard deviation is $\left({\mu }_x\right)$.
The standard deviation $\left(\sigma \right)$ is assumed to be $1.1$ hours.
$\begin{array}{r}{\sigma }_X=\frac{\sigma }{\sqrt{n}}\end{array}$

$=\frac{1.1}{\sqrt{40}}$

$=\frac{1.1}{6.3246}$
$=0.17$

As a result, the mean and standard deviation of all possible sample spans of binge drinking episodes for size $40$ binge drinkers are $4.9$ hours and $0.17$ hours, respectively.

Part (a) Step 3: Given information

To find the mean and standard deviation of all possible sample mean spans of binge drinking episodes for sample size $n=120$.

Part (b) Step 4: Explanation

Let, the Mean is $4.9$ hours and the Standard deviation is$1.1$hours.

Determine the sample size $120$ for the mean $\left({\mu }_X\right)$ as:
${\mu }_x=\mu$
${\mu }_x=44.8$
Then the standard deviation $\left({\sigma }_X\right)$can be determine as:

$\begin{array}{r}{\sigma }_X=\frac{\sigma }{\sqrt{n}}\end{array}$
$=\frac{1.1}{\sqrt{120}}$
$=\frac{1.1}{6.3246}$
$=0.1004$

As a result, the mean and standard deviation of all potential sample spans of binge drinking episodes for samples of size $120$ binge drinkers are $4.9$ hours and $0.1004$ hours, respectively.