Question

A study on students drinking habits wants to determine the true average number of alcoholic drinks all fsu undergraduate students who are members of a fraternity or sorority have in a one week period. We know from preliminary studies that the standard deviation is around 2.6. How many students should be sampled to be within 0.25 drinks of population mean with 95% probability?

Answer 1

Answer:

416

Step-by-step explanation:

Standard deviation is 2.6

Margin error for the problem is 0.25

Probability 95%, that means thet the siginficance level α is 1 – p

α = 1 – 0.95 = 0.05

margin of error (ME) can be defined as follows

ME = Z(α/2) * standard deviation/ √n

Where n is the sample size

Z(0.05/2) = Z(0.025)

Using a z table Z = 1.96

Now, replacing in the equation and find n

.25 = 1.96 * 2.6/ √n

0.25 = 5.096/√n

√n = 5.096 /0.25

√n = 20.38

n = 20.38^2

n = 415.5 = 416