Given
[tex]H0: \ \mu=45 \\ \\ H1: \ \mu\ \textless \ 45[/tex]
This is a one-tailed test.
[tex]z= \frac{41.8-45}{6/ \sqrt{25} } = \frac{-3.2}{6/5} = \frac{-3.2}{1.2} =-2.6667[/tex]
[tex]P(z\ \textless \ -2.6667)=0.00383[/tex]
Since the p-value of the sample statistic (0.00383) is less that the significant level (0.025), we reject the null hypothesis.