<h2>
Answer with explanation:</h2>
Let
be the population mean lifetime of circulated $1 bills.
By considering the given information , we have :-

Since the alternative hypotheses is two tailed so the test is a two tailed test.
We assume that the lifetime of circulated $1 bills is normally distributed.
Given : Sample size : n=50 , which is greater than 30 .
It means the sample is large so we use z-test.
Sample mean : 
Standard deviation : 
Test statistic for population mean :-


The p-value= 
Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.
Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.