Question

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". 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.

Part (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.

Part (b): Repeat part (a) with $n=120$.

Short answer

Part (a): The mean and standard deviation of are 4.9 hours and 0.17 hours.

On interpreting, for samples of size 40 binge drinkers, the mean and standard deviation of all possible sample spans of binge drinking episodes are 4.9 hours and 0.17 hours, respectively.

Part (b): The mean and standard deviation of are 44.8 hours and 0.1004 hours.

On interpreting, for samples of size 120 binge drinkers, the mean and standard deviation of all possible sample spans of binge drinking episodes are 44.8 hours and0.1004 hours, respectively.

Step-by-step solution

Part (a) Step 1.  Given information.

Consider the given question,

The mean $\left(\mu \right)$ span of binge drinking episodes is 4.9 hours.

The standard deviation is 1.1hours.

Part (a) Step 2. Find the mean and standard deviation when n=40.

The mean span of binge drinking episodes is 4.9 hours.

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =4.9 \end{gathered}$$

Thus, the mean $\left({\mu }_{\overline{x}}\right)$ of all possible sample mean spans of binge drinking episodes for sample size 40is $4.9$ hours.

The standard deviation $\left({\sigma }_{\overline{x}}\right)$ of all possible sample mean spans of binge drinking episodes for sample size 40.

It is given that the standard deviation $\left(\sigma \right)$ to be 1.1hours.

$$\begin{gathered} {\overline{\sigma }}_x=\frac{\sigma }{\sqrt{n}} \\ =\frac{1.1}{\sqrt{40}} \\ =\frac{1.1}{6.3246} \\ =0.17 \end{gathered}$$

Thus, the standard deviation $\left({\sigma }_{\overline{x}}\right)$of all possible sample mean spans of binge drinking episodes for sample size $40$is $0.17$hours.

On interpreting, we can say that for samples of size 40 binge drinkers, the mean and standard deviation of all possible sample spans of binge drinking episodes are4.9 hours and0.17 hours, respectively.

Part (b) Step 1. Find the mean and standard deviation when n=120.

The mean $\left({\mu }_{\overline{x}}\right)$ of all possible sample mean spans of binge drinking episodes for sample size is 120 hours.

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =44.8 \end{gathered}$$

Thus, the mean $\left({\mu }_{\overline{x}}\right)$ of all possible sample mean spans of binge drinking episodes for sample size 120 is 44.8 hours.

It is given that the standard deviation $\left(\sigma \right)$ of all possible sample mean spans of binge drinking episodes for sample size 40.

role="math" localid="1652624273272" $$\begin{gathered} {\overline{\sigma }}_x=\frac{\sigma }{\sqrt{n}} \\ =\frac{1.1}{\sqrt{120}} \\ =\frac{1.1}{10.9544} \\ =0.1004 \end{gathered}$$

Thus, the standard deviation $\left({\sigma }_{\overline{x}}\right)$ of all possible sample mean spans of binge drinking episodes for sample size 120 is 0.1004 hours.

On interpreting, we can say that for samples of size 120 binge drinkers, the mean and standard deviation of all possible sample spans of binge drinking episodes are 44.8 hours and0.1004 hours, respectively.