Question

A random sample of 11fields of spring wheat has a mean yield of 20.2bushels per acre and standard deviation of 5.19 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

Answer 1

Answer:

1. $t_{0.01/2,(10)}=3.169$

2. Confidence interval = [15.2, 25.2]

Step-by-step explanation:

X = 20.2

S = 5.19

n = 11

Step 1 of 2: Critical value

α = 1 - 0.99 = 0.01

df = 11 - 1 = 10

Looking at t distribution table with α = 0.01 and df = 10, we find

$t_{0.01/2,(10)}=3.169$

Step 2 of 2: 99% confidence interval

$std-err=\frac{S}{\sqrt{n}}=\frac{5.19}{\sqrt{11}}=1.5648$

$X+t_{0.01/2,10}*std-err$

20.2 + 3.169*1.5648

20.2 + 4.9595

25.2

$X-t_{0.01/2,10}*std-err$

20.2 - 3.169*1.5648

20.2 - 4.9595

15.2

Confidence interval = [15.2, 25.2]

Hope this helps!