Question

An engineer has carefully measured the spacing between anchor bolts (in mm) as:

19   23    21    21
19   13    26    15
30    30   17    24
20   16    24    20

Find:
a. The 90 percent CI for µ
b. The 80 percent CI for µ
c. The 99 percent CI for µ assuming σ = 4.1 mm is known

Answer 1

= (19 + 23 + 21 + 21 + 19 + 13 + 26 + 15 + 30 + 30 + 17 + 24 + 20 + 16 + 24 + 20)/16 = 21.75

S = sqrt(((19 - 21.75)^2 + (23 - 21.75)^2 + (21 - 21.75)^2 + (21 - 21.75)^2 + (19 - 21.75)^2 + (13 - 21.75)^2 + (26 - 21.75)^2 + (15 - 21.75)^2 + (30 - 21.75)^2 + (30 - 21.75)^2 + (17 - 21.75)^2 + (24 - 21.75)^2 + (20 - 21.75)^2 + (16 - 21.75)^2 + (24 - 21.75)^2 + (20 - 21.75)^2)/15) = 5.2345

A) At 90% confidence interval the critical value is t^* = 1.753

The 90% confidence interval for is

+/- t^* * s/

= 21.75 +/- 1.753 * 5.2345/

= 21.75 +/- 2.294

= 19.456, 24.044

B) At 80% confidence interval the critical value is t^* = 1.341

The 80% confidence interval for is

+/- t^* * s/

= 21.75 +/- 1.341 * 5.2345/

= 21.75 +/- 1.755

= 19.995, 23.505

C) At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for is

+/- z_0.005 *

= 21.75 +/- 2.58 * 4.1/

= 21.75 +/- 2.6445

= 19.1055, 24.3945