Assume a recent sociological report states that university students drink 5.70 alcoholic drinks per week on average, with a standard deviation of 1.7201. Supposed Jason, a policy manager at a local university, decides to take a random sample of 170 university students to survey them about their drinking habits. Determine the mean and standard deviation of the sampling distribution of the sample mean alcohol consumption. Provide your answer with precision to two decimal places. mean of the sampling distribution

The mean of the sampling distribution of the sample mean alcohol consumption is of 5.70 alcoholic drinks per week and the standard deviation is 0.13 alcoholic drinks per week

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 5.70, \sigma = 1.7201, n = 170, s = \frac{1.7201}{\sqrt{170}} = 0.13[/tex]

The mean of the sampling distribution of the sample mean alcohol consumption is of 5.70 alcoholic drinks per week and the standard deviation is 0.13 alcoholic drinks per week