Question

A farmer outside of Clinton NC plants 20 plots of a new type of corn. The average yield for these plots is ī= 150 bushels per acre. Assume that the yield
per acre for the new viety of corn follows a normal distribution with
unknown mean and standard deviation of 20 bushels. Find a 95% confidence
interval.
A. 20 + 3.2
B. 150 +2.5
C. 150 +3.2
D. 150 +4.3

Answer 1

Answer:

The 95% confidence  interval is $(150\pm 8.8)$.

Step-by-step explanation:

The information provided is:

$\bar x=150\\\sigma = 20\\n=20$

The critical value for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$

          $=1.96\times\frac{20}{\sqrt{20}}\\\\=1.96\times 4.47214\\\\=8.7653944\\\\\approx 8.8$

Then the 95% confidence  interval is:

$CI=\bar x\pm MOE$

     $=150\pm 8.8$

Thus, the 95% confidence  interval is $(150\pm 8.8)$.