Question

A random sample of 10 fields of corn has a mean yield of 22.5 bushels per acre and standard deviation of 5.84 bushels per acre. Determine the 99% 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

The critical value that should be used in constructing the confidence interval is T = 3.25.

The 99% confidence interval for the true mean yield is between 16.498 and 28.502 bushels per acre.

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 10 - 1 = 9

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.99)/(2) = 0.995`. So we have T = 3.25 which is the critical value that should be used in constructing the confidence interval.

The margin of error is:

`M = T(s)/(√(n)) = 3.25(5.84)/(√(10)) = 6.002`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 22.5 - 6.002 = 16.498 bushels per acre

The upper end of the interval is the sample mean added to M. So it is 22.5 + 6.002 = 28.502 bushels per acre

The 99% confidence interval for the true mean yield is between 16.498 and 28.502 bushels per acre.