A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standa

Question

2 years ago

14

rd deviation of 0.77 mpa.

1 answer:

nignag [31] · Answer 1 · 2022-08-07T01:08:47+03:00

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n

Where:

α= 5%

α/2 = 2.5%

t 0.025, 19 = 2.093 (check t table)

n = 20

df = n – 1 = 20 – 1 = 19

So plugging in our values:

8.41 ± 2.093 * 0.77 √ 1 + 1/20

= 8.41 ± 2.093 * 0.77 (1.0247)

= 8.41 ± 2.093 * 0.789019

= 8.41 ± 1.65141676