PLEASE DOUBLE CHECK ANSWER, AND NO HANDWRITTEN ANSWERS. THANK YOU!!
A random sample of 7 fields of durum wheat has a mean yield of 29.8 bushels per acre and standard deviation of 3.62 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.
Step 2 of 2: Construct the 99% confidence interval. Round your answer to one decimal place.
Solution :
Given that,
Point estimate = sample mean = = 29.8
sample standard deviation = s = 3.62
sample size = n = 7
Degrees of freedom = df = n - 1 = 6
t /2,df = t0.005,6 = 3.707
Margin of error = E = t/2,df * (s /n)
= 3.707 * (3.62 / 7)
Margin of error = E = 5.1
The 99% confidence interval estimate of the population mean is,
- E < < + E
29.8 - 5.1 < < 29.8 + 5.1
24.7 < < 34.9
(24.7 , 34.9)
Get Answers For Free
Most questions answered within 1 hours.