Question

A random sample of 9 fields of corn has a mean yield of 38.2 bushels per acre and standard deviation of 9.66 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

The critical value that should be used is T = 1.397.

The 80% confidence interval for the true mean yield is between 24.705 bushels per acre and 51.695 bushels per acre.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 9 - 1 = 8

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.8}{2} = 0.9$. So we have T = 1.397. This is the critical value that should be used.

The margin of error is:

M = T*s = 1.397*9.66 = 13.495

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 38.2 - 13.495 = 24.705 bushels per acre.

The upper end of the interval is the sample mean added to M. So it is 38.2 + 13.495 = 51.695 bushels per acre.

The 80% confidence interval for the true mean yield is between 24.705 bushels per acre and 51.695 bushels per acre.