Answer to Question #115482 in Statistics and Probability for Yow

Question

1. The recorded measurements of sample of soil cohesion are 96, 106, 97, 88, 102 and 94 KPa respectively. (a) Compute the sample mean cohesion. This is considered as the unbiased and efficient estimate of the population mean. (b) Compute the sample variance. (c) Compute the estimated population variance by multiplying the sample variance by (n/n-1). (d) Compare the sample standard deviation with the estimated population standard deviation.

Answer 1

(a) $\mu = (\sum^n xi)/n=(96+106+97+88+102+94)/6=97.166$

(b)VAR=$(\sum^n (xi-\mu)^2)/(n-1))=39.366$

(c)pVAR=VAR*(n-1)/n=32.8

(d)sample standard deviation= $\sqrt{VAR}=6.27$

population standard deviation=$\sqrt{pVAR}=5.727$

sample standard deviation>population standard deviation