Question

Historical data indicated that the time required to service the conveyor belt at Coca-Cola Amatil's Richlands operations can be modelled as a normal distribution with a standard deviation of 20 minutes. A random sample of 20 services revealed a mean service duration of 119.9 minutes. Determine a 90% confidence interval for the mean service time in minutes. State the lower bound of this interval correct to two decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 119.9

sample standard deviation = s = 20

sample size = n = 20

Degrees of freedom = df = n - 1 = 19

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,19 = 1.729

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

= 1.729 * ( 20/ 20)

= 7.73

The 90% confidence interval estimate of the population mean is,

- E < < + E

119.9 - 7.73 < < 119.9 + 7.73

112.17 < < 127.63

(112.17 , 127.63)

The lower bound of this interval is 112.17.