Final answer:
To determine the 99% confidence interval for the true mean yield of a random sample of 7 fields of durum wheat, we need to find the critical value using a t-distribution table or calculator. The critical value is approximately 3.707 for a 99% confidence level with 6 degrees of freedom. Using the sample mean, sample standard deviation, and sample size, we can calculate the confidence interval as (20.242, 34.558).
Step-by-step explanation:
To determine the 99% confidence interval for the true mean yield of a random sample of 7 fields of durum wheat, we need to find the critical value. Since the sample size is small (less than 30), we use a t-distribution. The formula to find the critical value is x-bar ± t * (s / √n), where x-bar is the sample mean, t is the critical value, s is the sample standard deviation, and n is the sample size.
First, we need to find the critical value by using a t-distribution table or a calculator. For a 99% confidence level with 6 degrees of freedom (n-1), the critical value is approximately 3.707.
Next, we can calculate the confidence interval: 27.4 ± 3.707 * (5.75 / √7). This gives us a 99% confidence interval of (20.242, 34.558) for the true mean yield of durum wheat in the population.
The Incomplete Question:
A random sample of 7 fields of durum wheat has a mean yield of 27.4 bushels per acre and standard deviation of 5.75 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