Question

A random sample of 18 fields of barley has a mean yield of 36.4 bushels per acre and standard deviation of 7.37 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is approximately normal.

Answer 1

Answer:

So, the confidence interval is [34.08, 38.72].

Step-by-step explanation:

We know that a  random sample of 18 fields of barley has a mean yield of 36.4 bushels per acre and standard deviation of 7.37 bushels per acre.  

We have :

$n=18\\\\\mu=36.4\\\\\sigma=7.37\\\\c=80\%$

We use the Appendix: Critical Values Tables, and we get that:

$t_{\frac{\alpha}{2}}=1.333$

We calculate the margin of error:

$E=t_{\frac{\alpha}{2}}\cdot \frac{\sigma}{\sqrt{n}}=1.333\cdot \frac{7.37}{\sqrt{18}}=2.32$

We get the boundaries of the confidence interval:

$\mu-E=36.4-2.32=34.08\\\\\mu+E=36.4+2.32=38.72$

So, the confidence interval is [34.08, 38.72].