Answer to Question #115323 in Statistics and Probability for Angelica

Question

The recorded measurements of sample of soil cohesion are 90, 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) sample mean

$\bar x =\frac{{\sum_{i=1}^n} x_i} {n}$

$=\frac{90+106+97+88+102+94}{6}$

=96.1667

b) sample variance

$S^2=\frac {\sum_{i=1}^n( {x_i-\bar x})^2 } {n-1}$

$=\frac {(90-96.1667)^2+(106-96.1667) ^2+...+(94-96.1667)^2}{6-1}$ =48.1667

c) estimated population variance

$\sigma ^2 =\frac{n-1} {n}s^2$

=$\frac{5}{6}*48.1667$

=40.138

d) The estimated population Variance is lower than the sample Variance.