Simple calculating confidence interval for $u$ An agricultural researcher plants in a random sample of $25$ plots with a new variety of corn. The average yield for these plots is $\bar{x} = 150$ bushels per acre with a standard deviation $s = 10$ bushels per acre . Assume that the yield per acre for the new variety of corn follows a Normal distribution with unknown mean µ and standard deviation σ. Calculate A $95\%$ confidence interval for $u$ **my attempt** $n = 25, \bar{x} = 150, s = 10, a = 0.95$. Keyword: Unknown mean $u$ means we use $t$ table. $$\bar{x} \pm t_{\frac{1+0.95}{2}} \frac{s}{\sqrt{n}}$$ $$150 \pm t_{0.975} \frac{10}{\sqrt{25}}$$ $$150 \pm t_{0.975} \times 2$$ But this is wrong according to the solution: $$150 \pm t_{0.95} \times 2$$ How?

I agree with you. The confidence interval is

$$\large{\left[\overline x-t_{(1-\tfrac{\alpha}{2};n-1)}\cdot \frac{s}{\sqrt n} ; \ \overline x+t_{(1-\tfrac{\alpha}{2};n-1)}\cdot \frac{s}{\sqrt n} \right]}$$

In your case $1-\tfrac{\alpha}{2}=1-\frac{0.05}{2}=1-0.025=0.975, n=25, \overline x=150$ and $s=10$. Thus the interval is

$\left[150-t_{(0.975;24)}\cdot \frac{10}{5} ; \ 150+t_{(0.975;24)}\cdot \frac{10}{5} \right]$, where $t_{(0.975;24)}=2.064$