Question

In a survey, 23 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $49 and standard deviation of $8. Find the margin of error at a 98% confidence level.

Answer 1

Solution:

Given that,

n = 23

= 49

s = 8  

Note that, Population standard deviation() is unknown..So we use t distribution.

Our aim is to construct 98% confidence interval.   

c = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.022 = 0.01

Also, d.f = n - 1 = 23 - 1 = 22

    =    =  _{0.01 , 22} = 2.508

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n)

= 2.508 * (8 / 23)

= 4.1836

Margin of error E = 4.1836