Let [tex]\mu[/tex] be the population mean lifetime of circulated $1 bills.
By considering the given information , we have :-
[tex]H_0:\mu=18\\\\H_a:\mu\neq18[/tex]
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 : [tex]\overline{x}=18.8[/tex]
Standard deviation : [tex]\sigma=2.8[/tex]
Test statistic for population mean :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02[/tex]
The p-value= [tex]2P(z>2.02)=0.0433834[/tex]
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.