Question

The article “Ultimate Local Capacities of Expansion Anchor Bolts” (J. of Energy Engr., 1993: 139- 158) gave the following summary data on shear strength (kip) for a sample of 3/8-in anchor bolts: n= 18, mean is 4.25, and a sample standard deviation of 1.3. Calculate a 95% confidence interval for true average shear strength. Interpret your finding.

Answer 1

Solution :

Given that,

= 4.25

s =1.3

n = Degrees of freedom = df = n - 1 =18 - 1 = 17

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,17 = 2.110 ( using student t table)

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

= 2.110 * (1.3 / 18)

= 0.6

The 95% confidence interval mean is,

- E < < + E

4.25 - 0.6 < < 4.25+ 0.6

3.65 < < 4.85