Question

A random sample of 80 Mizzou students showed that 20 had driven a car during the day before the survey was conducted. Suppose that we are interested in forming a 68 percent confidence interval for the proportion of all Mizzou students who drove a car the day before the survey was conducted. Where appropriate, express your answer as a proportion (not a percentage). Round answers to three decimal places.

(a) The estimate is: .

(b) The standard error is:

(c) The multiplier is:

Answer 1

Solution :

Given that,

n = 80

x = 20

(a)

= x / n = 20 / 80 = 0.250

1 - = 1 - 0.250 = 0.750

(b)

Standard error =  (( * (1 - )) / n)

= (0.25 * 0.75) / 80

= 0.048

Standard error = 0.048

(c)

At 68% confidence level the z is ,

= 1 - 68% = 1 - 0.68 = 0.32

/ 2 = 0.32 / 2 = 0.16

Z_/2 = Z_0.16 = 0.99

Multiplier = 0.99