Question

A random sample of ten metal rods produced by a machine is taken. Each rod is tested for
hardness. The results, in suitable units, are as follows
525 520 522 524 518 520 519 525 527 516
Calculate an unbiased estimate of the population variance.
Find a 98% confidence interval for the population mean.

Answer 1

n=10, x ={525, 520, 522, 524, 518, 520, 519, 525, 527, 516 }

$\bar{x}=\frac{\displaystyle\sum_{i=1}^n x_i}{n}$ = (525+520+522+524+518+520+519+525+527+516)/10 = 521.6  

An unbiased estimate of the population variance.  

$\sigma^{2}=\frac{\displaystyle\sum_{i=1}^n (x_i -\bar{x})^{2}}{n-1} =\\= ((525-521.6)^2+(520-521.6)^2+(522-521.6)^2+...+(516-521.6)^2)/(10-1) = 12.71$

A 98% confidence interval for the population mean:

$\bar{x}-\Delta_x \le\bar{x}\le\bar{x}+\Delta_x$

$∆ _x= t *l,$

Find t in the t-Table : t(9,0.02) = 2.82,

$l=\sqrt{\smash[b]{\sigma^{2}/n}} =\sqrt{ 12.71/10}=$ 1.1274,

$\Delta_x$ = 1.1274*2.82 = 3.1793 ,

521.6- 3.1793 $\le\bar{x}\le$ 521.6 +3.1793

518.42 $\le\bar{x}\le$ 524.78