Answer: Option 'b' is correct.
Step-by-step explanation:
Since we have given that
Average age = 25 years
Standard deviation = 2 years
Sample size = 16
So, Null hypothesis : [tex]H_0:\mu\leq 24[/tex]
Alternate hypothesis: [tex]H_1:\mu>24[/tex]
Level of significance = 5% = 0.05
So, test statistic would be
[tex]\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\=\dfrac{25-24}{\dfrac{2}{\sqrt{16}}}\\\\=\dfrac{1}{\dfrac{2}{4}}}\\\\=\dfrac{1}{0.5}\\\\=2[/tex]
Hence, Option 'b' is correct.