For a 90 % confidence interval and 17 students, the t-value for this calculation will be

$$ mean:\;t\left(\alpha, N\right)=t\left(.05,17\right)=1.740 $$

$$ stddev:\;t\left(\alpha, N-1\right)=t\left(.05,16\right)=1.746 $$

The 90 % confidence interval for mean:

$$ \varDelta =t\left(\alpha, N\right)\bullet \frac{\sigma_N}{\sqrt{N}}=1.740\bullet \frac{0.6}{\sqrt{17}}=0.253 $$

$$ P\left(\mu -\varDelta \le {\mu}_{TRUE}\le \mu +\varDelta \right)=90\% $$

$$ P\left(2.5-.253\le {\mu}_{TRUE}\le 2.5+.253\right)=90\% $$

$$ P\left(2.247\le {\mu}_{TRUE}\le 2.753\right)=90\% $$

The 90 % confidence interval for standard deviation:

$$ \varDelta =t\left(\alpha, N-1\right)\bullet \frac{\sigma_N}{\sqrt{2N}}=1.746\bullet \frac{0.6}{\sqrt{2\bullet 17}}=0.18 $$

$$ P\left(\sigma -\varDelta \le {\sigma}_{TRUE}\le \sigma +\varDelta \right)=90\% $$

$$ P\left(0.6-.18\le {\sigma}_{TRUE}\le 0.6+.18\right)=90\% $$

$$ P\left(0.42\le {\sigma}_{TRUE}\le 0.78\right)=90\% $$

For a 95 % confidence interval and 17 students, the t-value for this calculation will be

$$ mean:\;t\left(\alpha, N\right)=t\left(.025,17\right)=2.11 $$

$$ stddev:\;t\left(\alpha, N-1\right)=t\left(.025,16\right)=2.12 $$

The 95 % confidence interval for mean:

$$ \varDelta =t\left(\alpha, N\right)\bullet \frac{\sigma_N}{\sqrt{N}}=2.11\bullet \frac{0.6}{\sqrt{17}}=0.307 $$

$$ P\left(\mu -\varDelta \le {\mu}_{TRUE}\le \mu +\varDelta \right)=95\% $$

$$ P\left(2.5-.307\le {\mu}_{TRUE}\le 2.5+.307\right)=95\% $$

$$ P\left(2.193\le {\mu}_{TRUE}\le 2.807\right)=95\% $$

The 95 % confidence interval for standard deviation:

$$ \varDelta =t\left(\alpha, N-1\right)\bullet \frac{\sigma_N}{\sqrt{2N}}=2.12\bullet \frac{0.6}{\sqrt{2\bullet 17}}=.218 $$

$$ P\left(\sigma -\varDelta \le {\sigma}_{TRUE}\le \sigma +\varDelta \right)=95\% $$

$$ P\left(0.6-.218\le {\sigma}_{TRUE}\le 0.6+.218\right)=95\% $$

$$ P\left(0.382\le {\sigma}_{TRUE}\le 0.818\right)=95\% $$