Answer: We reject the null hypothesis.
Step-by-step explanation:
Since we have given that
Null and alternate hypothesis are
[tex]H_0:\mu =30\\\\H_1:\mu \neq 30[/tex]
n = 25 and α = 0.05
Degrees of freedom would be
[tex]n-1\\\\=25-1\\\\=24[/tex]
Using t-table, we get that
[tex]t_{0.05,24}=2.0638[/tex]
[tex]\bar{X}=29.5\\\\s=1.8\\\\n=25[/tex]
[tex]t=\dfrac{\bar{X}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\t=\dfrac{29.5-30}{\dfrac{1.8}{\sqrt{25}}}\\\\t=\dfrac{-0.5}{0.36}\\\\t=-1.39[/tex]
Since -1.39<2.0638
So, we reject the null hypothesis.