Assume a recent sociological report states that university students drink 5.10 alcoholic drinks per week on average, with a standard deviation of 1.3401. Suppose Jason, a policy manager at a local university, decides to take a random sample of 150 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.

a. mean of the sampling distribution =
b. standard deviation of the sampling distribution =

Respuesta :

Answer:

a) 1.34

b) 0.11

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.

Population:

Mean is 5.10, standard deviation is 1.34

a. mean of the sampling distribution =

By the Central Limit Theorem, the mean of the sampling distribution is 5.1

b. standard deviation of the sampling distribution =

n = 150

By the Central Limit Theorem

[tex]s = \frac{1.3401}{\sqrt{150}} = 0.11[/tex]