Question

Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a major university. They found that students in the sample spent an average of 2.3 hours per day on social media and an average of 1.8 hours per day on studying. If all the students at the university in fact spent 2.2 hours per day on studying, with a standard deviation of 2 hours, and we find the probability of observing a sample mean of 1.8 hours studying has an extremely low probability, we say that observed time is:

Answer 1

Answer:

Consider the following calculations

Step-by-step explanation:

Given that,

mean = u = 2.2

standard deviation = sigma = 2

n = 200

The sampling distribution of mean and standard deviation is ,

Ux = 2.2 and

U x = sigma /  n = 2 / \sqrt 200 = 0.141

N(2.2 , 0.141)

Answer 2

Using the normal distribution and the central limit theorem, it is found that the observed time is: N(2.2, 0.1414).

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• The mean is of $\mu = 2.2$.

• The standard deviation is of $\sigma = 2$.

• A sample of 200 is taken, hence $n = 200, s = \frac{2}{\sqrt{200}} = 0.1414$.

Hence, by the Central Limit Theorem, the observed time is: N(2.2, 0.1414).

To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213