Question

In a survey, 26 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $41 and standard deviation of $2. Find the margin of error at a 80% confidence level.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 41

sample standard deviation = s = 2

sample size = n = 26

Degrees of freedom = df = n - 1 = 25

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t /2,df = t0.10,25 = 1.316

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

= 1.316 * (2 / 26)

= 0.516

The margin of error at a 80% confidence level is 0.516.