Question

The Student Recreation Center wanted to determine what sort of physical activity was preferred by students. In a survey of 58 students, 41 indicated that they preferred outdoor exercise over exercising in a gym. When estimating the proportion of all students at the university who prefer outdoor exercise with 99% confidence, what is the margin of error?

Question 8 options:

1)

0.1542

2)

0.0202

3)

0.0598

4)

0.0092

5)

0.1393

Answer 1

Solution :

Given that,

n = 58

x = 41

Point estimate = sample proportion = = x / n = 41/580.707

1 - = 1-0.707=0.293

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.58

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 (((0.707*0.293) /58 )

=0.1542

correct option 1