A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.2 hours. The

researcher feels that college students do not sleep 7.2 hours on average. The researcher randomly selected 15 students and

found that on average they slept 8.3 hours. The standard deviation of the sample is 1.2 hours. If you wanted to test this claim,

what type of statistical test would you do and why?

Calculate the value of z and its probability, since with this we can know how likely it is that this will happen.

We have that the mean (m) is equal to 8.3, the standard deviation (sd) 1.2 and the sample size (n) = 15

They ask us for P (x =7.2)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (7.2 - 8.3) / (1.2 / (15 ^ (1/2))

z = -3.55

With the normal distribution table (attached), we have that at that value the approximate probability is:

P (z = -3.55) = 0.0001

The probability is 0.01 %

This affirms that the students do not sleep what the study says because the probability of this happening according to the survey is almost nil.

step-by-step explanation:

if you want to solve percent error follow these steps.

1. subtract the accepted value from the experimental value.

2. take the absolute value of step 1.

3. divide that answer by the accepted value.

4. multiply that answer by 100 and add the % symbol to express the answer as a percentage.

if you want to solve percent of change just do this.

1. work out the difference (increase) between the two numbers you are comparing.

2. divide the increase by the original number and multiply the answer by 100. if your answer is a negative number then this is a percentage decrease.

hope this : )

3,750 deer

step-by-step explanation:

1,500 is 15 times larger than 100

lets make 250 15 times larger.

3750 deer