Question

3. A sample of 30 randomly selected TV sets has a repair cost mean of $ 356.25, with a standard deviation of $ $19.75. Find the margin of error E for a 95% confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $356.25

sample standard deviation = s = $19.75

sample size = n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,29 = 2.045

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

= 2.045 * (19.75 / 30)

Margin of error = E = 7.37