Question

An analyst at Carnival Cruise Lines is interested in the time it takes for a room service order to go from time of order to time of delivery. In order to do this, she takes a sample of 49 random guests and finds that their average time to delivery was 11.2 minutes and that the sample standard deviation was 3.5 minutes. Calculate a 95% Confidence Interval around the mean.

Answer 1

Solution :

Given that,

= 11.2

s =3.5

n =49

Degrees of freedom = df = n - 1 = 49- 1 = 48

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,48 =2.011 ( using student t table)

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

= 2.011* (3.5 / 49)

= 1.01

The 95% confidence interval estimate of the population mean is,

- E < < + E

11.2 - 1.01 < <11.2 + 1.01

10.19 < < 12.21

(10.19 , 12.21)