Question

When people smoke, the nicotine they absorb is converted to nicotine, which can be measured. A sample of 40 smokers has a mean nicotine level of 172.5. Assuming that σ is known to be 119.5, a 90% confidence interval estimate for the mean nicotine level is [141.4, 203.6]. How large a sample would you need to reduce its length to 50?

Answer 1

Z(0.9)= 1.645 from z table or =NORM.S.INV(0.95) in Excel formula.

Marginal error is given by

$E=\frac{Z(0.9)×\sigma}{\sqrt{n}}$

To get a length of 50, E=25.

$25=\frac{1.645×119.5}{\sqrt{n}}$

$\sqrt{n}=\frac{196.56}{25}, n=7.8624^2=61.82$

The sample size should be 62 for the length of the confidence interval to be reduced to 50.