According to the National Institute on Alcohol Abuse and Alcoholism (NIAAA), and the National Institutes of Health (NIH), 41% of college students nationwide engage in "binge drinking" behavior, having five or more drinks on one occasion during the past two weeks. A college president wonders if the proportion of students enrolled at her college that binge drink is lower than the national proportion. In a commissioned study, 462 students are selected randomly from a list of all students enrolled at the college. Of these, 162 admitted to having engaged in binge drinking.

Required:

Using the most conservative estimate, what sample size would be required to reduce the margin of error to 2 percentage points?

The sample size required is 2188.

Step-by-step explanation:

The (1 - α)% confidence interval for the population proportion is:

The margin of error for this interval is:

The information provided is:

X = 162

n = 462

MOE = 0.02

Assume the confidence level as 95%.

Compute the sample proportion as follows:

The z-critical value for 95% confidence interval is:

Compute the sample size required as follows:

Thus, the sample size required is 2188.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.