In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking websites from their college library. what is the margin of error for the true proportion of all undergraduates who access social networking sites from their college library? hurry

Margin of Error = 9.01%

Explanation:

Given

Sample size, n = 50

Number of regular access = 44

The level of confidence is not stated.

I'll assume a 95% level of confidence, so z* = 1.96

To calculate Margin of error,

First we need to calculate the proportion of those that have regular access and those that don't, as probabilities.

Let P = Probability of those with regular access.

P = 44/50 = 0.88

Let Q = probability of those that don't have regular access

P + Q = 1 --- make Q the subject of formula

Q = 1 - P

Q = 1 - 0.88

Q = 0.12

First, we calculate the standard error

Standard Erro is calculated using the following formula.

A.E = √(PQ/n)

By substituton

S.E = √(0.88*0.12/50)

S.E = 0.045956501172304

Finally,. margin of error is calculated by S.E * z*

= 0.045956501172304 * 1.96

Margin of Error = 0.090074742297716

Margin of Error = 9.01% --- Approximated

i think 5 miles per hour

step-by-step explanation: