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.

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37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3