We are given the following data
Significance level = 0.1
For the first data
[tex]\begin{gathered} n=45 \\ \bar{x_1}=3.47 \\ \sigma_1=0.07 \end{gathered}[/tex]
For the second data
[tex]\begin{gathered} n=45 \\ \bar{x}_2=3.49 \\ \sigma_2=0.02 \end{gathered}[/tex]
We can get the test statistic as follow
[tex]z=\frac{\bar{x_1-}\bar{x_2}}{\sqrt[]{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}}[/tex]
Thus
[tex]\begin{gathered} z=\frac{3.47-3.49}{\sqrt[]{\frac{0.07^2}{45}+\frac{0.02^2}{45}}}=\frac{-0.02}{0.01085}=-1.842 \\ z=-1.84 \end{gathered}[/tex]
The test statistic is the z-score above = -1.84
The p-value of the z-score value obtained above is found out to be 0.07
Since the p-value obtained is less than the significance level, we will reject the null hypothesis
