Answer:
Suppose you want to assess student attitudes about the new campus center by surveying 100 students at your school. In this example, the group of 100 students represents the Sample, and all of the students at your school represent the Population.
Step-by-step explanation:
Previous concepts
The term sample represent a set of observations or individuals selected from a population. And we can have a random sample (when all the individuals have the same probability of being selected) or a non random sample (when not all the individuals have probability of inclusion into the sample)
The term population represent the total of observations or individuals with a common characteristic.
If N represent the sample of the population and n the sample size we have always this inequality:
Solution to the problem
Suppose you want to assess student attitudes about the new campus center by surveying 100 students at your school. In this example, the group of 100 students represents the Sample, and all of the students at your school represent the Population.
Answer:
2j -20 ≤ 46
Step-by-step explanation:
"20 less than double the number Jason [hit]" can be represented by 2j-20.
We are told that 46 is no fewer than this value, so it is greater than or equal to this value:
46 ≥ 2j -20
2j -20 ≤ 46 . . . . . swapping sides to match answer choices
The last equation would best describe the line.
y-4=3(x+2)
y-4=3x + 6
y=3x + 10
Answer:
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549
Step-by-step explanation:
We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of
). So we have T = 2.0930
The margin of error is:
M = T*s = 2.0930*0.16 = 0.3349.
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.92 - 0.3349 = 2.5851
The upper end of the interval is the sample mean added to M. So it is 2.92 + 0.3349 = 3.2549
The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549