We assume that the population distribution is approximately normal.
Let [tex]\mu[/tex] be the population mean.
By considering the given information , we have
[tex]H_0:\mu=5.4\\\\H_a:\mu>5.4[/tex], since the alternative hypothesis is right tailed , so the test is right tail test.
Given : Sample size : n=9, which is a small sample (n<30), so we use t-test.
Sample mean : [tex]\overline{x}=5.7[/tex]
Variance =[tex]\sigma^2=0.81[/tex]
Then, Standard deviation : [tex]\sigma=0.9[/tex]
Test statistic :-
[tex]t=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
i.e. [tex]t=\dfrac{5.7-5.4}{\dfrac{0.9}{\sqrt{9}}}=1[/tex]
Critical value of t=[tex]t_{9-1,0.025}=t_{8,0.025}=2.306[/tex]
Since the observed t-value is less than the critical value (1<2.306), so we do not reject the null hypothesis.
Thus , we conclude that we have evidence to do not reject the null hypothesis.
