Question

In a certain design, it is important that standard deviation of the current be less than 25 mA. In a test of n=18 parts, the sample standard deviation is found to be 20 mA. Can we be 95% confident that the standard deviation will not exceed 24 mA?

Answer 1

we have given,

sample standard deviation= 20 mA.

sample variance = 400mA

and n= 18

The critical values for α=0.05 and df = 17 degrees of freedom are χ_L²​=χ_1−α/2,₁₇​ =7.5642 and χ_U²​=χ_α/2,₁₇ =30.191 (from chi square table )

95% confidence interval for variance = $(\frac{(n-1)s^{2}}{\chi ^{2}_{\alpha /2,17}} ,\frac{(n-1)s^{2}}{\chi ^{2}_{1-\alpha /2,17}})$

= $(\frac{17*400}{7.5642} ,\frac{17*400}{30.191})$ = (225.2326,898.9731)

95% confidence interval for standard deviation = ($\sqrt{225.2326}%u200B,$$,\sqrt{898.9731}$​)=(15.0078,29.9829)

N0 ,We Cannot be 95% confident that the standard deviation will not exceed 24 mA