In: Statistics and Probability
A survey of 170 commuters shows that on average people travel 17.4 km with standard deviation 8.4 km. Find 99% confidence interval for the population mean commuting distance. State your conclusion.
Solution :
Given that,
Point estimate = sample mean = 
= 17.4
Population standard deviation =   
= 8.4
Sample size = n =170
At 99% confidence level the z is ,
  = 1 - 99% = 1 - 0.99 =
0.01
/ 2 = 0.01 / 2 = 0.005
Z
/2
= Z0.005 = 2.576   ( Using z table )
Margin of error = E = Z
/2*
(
/n)
= 2.576* (8.4 / 
170)
= 1.66
At 99% confidence interval is,
-
E < 
 < 
 + E
17.4-1.66 < 
 < 17.4+1.66
15.74< 
 < 19.06
(15.74 , 19.06 )
lower limit 15.74
lower limit 19.06
sample mean lies between these confidence interval