Question

An article presents measures of penetration resistance for a certain fine-grained soil. Fifteen measurements, expressed as a multiple of a standard quantity, had a mean of 2.64 and a standard deviation of 1.06. Find a 95% confidence interval for the mean penetration resistance for this soil. Round the answers to three decimal places.

Answer 1

Answer:

[2.053 , 3.227]

Step-by-step explanation:

The 95% confidence interval is given by the interval

$\large [\bar x-t^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]$

where

$\large \bar x$ = the sample mean  

s = the sample standard deviation  

n  = the sample size

$\large t^*$ is the 0.05 (5%) upper critical value for the Student's t-distribution with 14 degrees of freedom (sample size -1), which is an approximation to the Normal distribution for small samples (n<30).

Either by using a table or the computer, we find  

$\large t^*= 2.145$

and our 95% confidence interval is

$\large [2.64-2.145*\frac{1.06}{\sqrt{15}}, 2.64+2.145*\frac{1.06}{\sqrt{15}}]=\boxed{[2.053,3.227]}$