Question

The article "Influence of Penetration Rate on Penetrometer Resistance" (see textbook for reference) 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 1.02. Find a 95% of confidence interval for the mean penetration resistance for this soil.

Answer 1

Answer:

The 95% of confidence interval for the mean penetration resistance for this soil is (2.0823, 3.1977).

Step-by-step explanation:

The sample size is 15.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 15-1 = 14$

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 14 and 0.025 in the t-distribution table, we have $T = 2.145$.

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{1.02}{\sqrt{15}} = 0.26$

Now, we multiply T and s

$M = T*s = 2.145*0.26 = 0.5577$

For the lower end of the interval, we subtract the mean by M. So 2.64 - 0.5577 = 2.0823

For the upper end of the interval, we add the mean to M. So 2.64 + 0.5577 = 3.1977

The 95% of confidence interval for the mean penetration resistance for this soil is (2.0823, 3.1977).