Answer:
The 95% confidence interval for mean compressive strength is:
[tex]3237.3\leq \mu\leq 3282.5[/tex]
Step-by-step explanation:
We have a sample of n=12 specimens, with sample mean=3259.9 and sample standard deviation = 35.57.
We know the population is normally distributed.
For a 95% confidence interval and df=n-1=11, the t-statistic is t=2.201.
Then, the CI can be expressed as:
[tex]M-t*s/\sqrt{N}\leq \mu\leq M+t*s/\sqrt{N}\\\\3259.9-2.201*35.57/\sqrt{12}\leq \mu\leq 3259.9+2.201*35.57/\sqrt{12}\\\\ 3259.9-22.6\leq \mu\leq 3259.9+22.6\\\\3237.3\leq \mu\leq 3282.5[/tex]