Question

The an article gives 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

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)

= 0.647

The 95% confidence interval is,

- E < < + E

4.25 - 0.647 < < 4.25+ 0.647

3.603 < < 4.897

(3.603 , 4.897)

LOWER LIMIT 3.603

UPPER LIMIT 4.897