Question

In a survey, 101 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $36 and standard deviation of $6. Find the margin of error for a 99% confidence level.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $36

sample standard deviation = s = $6

sample size = n = 101

Degrees of freedom = df = n - 1 = 101-1=100

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01/ 2 = 0.005

t /2,df = t0.005,100 = 2.626

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

= 2.626 * (6 / 101)

E = 1.6

The margin of error for a 99% confidence level is 1.6