Question

In a survey, 21 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $4. Find the margin of error at a 95% confidence level. Give your answer to two decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $42

sample standard deviation = s = $4

sample size = n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,20 = 2.086

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

= 2.086 * (4 / 21)

Margin of error = E = 1.82