Question

A social sciences researcher wants to know about the amount of money that couples without children spend on leisure travel. Her sample of 28 couples yields a mean of $1263.00 and a standard deviation of $160.00. What is the margin of error for a 95% confidence interval?

Answer 1

Solution :

Given that,

= $1263

s =160

n = Degrees of freedom = df = n - 1 = 28- 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,90 = 2.052 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.052 * (160 / 28)

= 62.05

margin of error for a 95% confidence interva is 62.05