A confidence interval estimate is desired for the true mean

A confidence interval estimate is desired for the true mean gain in a circuit on a semiconductor device. Assume that the gain is normally distributed with a standard deviation = 20. How large must n be if the length of the 95% CI on mu is to be 50?

Solution

Answer to the question)

Standard deviation s =200

confidence level = 95% , this implies the value of Z critical = 1.96

The length of interval = 60

Length of interval = 2 * Margin of error (E)

this implies : 60 = 2 * Margin of error (E)

Margin of error = 60 /2 =30

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The formula of sample size is:

n = (Z*s / E)^2

.

We got all the values , on pluggng them we get

n = (1.96 *200 /30)^2

n = 170.7378 ~ 171

Thus the sample size must be 171 units