Question

the dean of student life at Gonzaga University reported recently that 0.75 of all Gonzaga students live on campus. Gus Spencer, a psychology professor at Gonzaga, has selected a random sample of 75 students for a behavior study.a. Calculate the standard error of the proportion. b. What is the probability the proportion of students in Gus’s study who live on campus is no greater than0.72?

Answer #1

Solution

Given that,

p = 0.75

1 - p = 1 - 0.75 = 0.25

n = 75

= p = 0.75

a) = [p ( 1 - p ) / n] = [(0.75 * 0.25) / 75 ] = 0.05

b) P( < 0.72)

= P[( - ) / < (0.72 - 0.75) / 0.05]

= P(z < -0.60)

Using z table,

= 0.2743

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