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**

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