Question

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation s = 25. (a) Find a 95% CI for m when n = 10 and x = 1000. (b) Find a 95% CI for m when n = 25 and x = 1000. (c) Find a 99% CI for m when n = 10 and x = 1000. (d) Find a 99% CI for m when n = 25 and x = 1000. (e) How does the length of the CIs computed change with the changes in sample size and confidence level?

uzing Z table please.

Answer 1

a) From the standard normal tables, we get:

P( -1.96 < Z < 1.96 ) = 0.95

Therefore the z value to be used here should be 1.96.

The confidence interval here thus is computed as:

This is the required 95% confidence interval here.

b) Here for n = 25, we get the confidence interval as:

This is the required 95% confidence interval here.

c) From the standard normal tables, we get:

P( -2.576 < Z < 2.576) = 0.99

Therefore the z value to be used here should be 2.576

The confidence interval here thus is computed as:

This is the required 95% confidence interval here.

d) Here we get the confidence interval as:

This is the required 95% confidence interval here.

e) From the above computations, we can clearly see that as the sample size increases, the margin of error decreases and therefore the confidence interval length decreases.

Also as the confidence level here increases, the z value increases and thus the confidence interval length also increases.