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"
"\u2206 _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
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