Question

A consumer took a random sample of 45 cups of coffee from a dispenser and found an average of 8.3 ounces and a standard deviation of 0.5 ounces. Compute the 95% margin of error of the sample mean. (Choose the option closest to the correct value, show work.)

a. 2.42 oz. b. 0.15 oz.                                c. 0.07 oz.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.3

sample standard deviation = s = 0.5

sample size = n = 45

Degrees of freedom = df = n - 1 = 45-1= 44

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,44 = 2.015

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

= 2.015 * (0.5 / 45)

= 0.15

95% Margin of Error = 0.15 oz

Option b is correct