Question

A national air traffic control system handled an average of 47,285 flights during 29 randomly selected days in a recent year. The standard deviation for this sample is 6,293 flights per day. The 99% confidence interval to estimate the average number of flights per day handled by the system is from

Question 7 options:

	a lower level of 54,223 to an upper level of 66,728.
	a lower level of 44,056 to an upper level of 50,514.
	a lower level of 34,562 to an upper level of 44,056.
	a lower level of 46,514 to an upper level of 54,568.

Answer 1

c )solution

Given that,

= 47285

s =6293

n = 29

Degrees of freedom = df = n - 1 =29 - 1 = 28

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,28 = 2.763    ( using student t table)

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

= 2.763 * ( 6293/ 29) = 3229

The 99% confidence interval is,

- E < < + E

47285 - 3229 < < 47285 + 3229

44,056  < < 50,514

a lower level of 44,056 to an upper level of 50,514