Question

A random sample of 10 fields of corn has a mean yield of 37.1 bushels per acre and standard deviation of 5.91 bushels per acre. Determine the 98% confidence interval for the true mean yield. Assume the population is approximately normal. Step 2 of 2: Construct the 98% confidence interval. Round your answer to one decimal place.

Answer 1

Solution :

Given that,

= 37.1

s =5.91

n = 10

Degrees of freedom = df = n - 1 =10 - 1 =9

At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,9 = 2.821 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.821 * ( 5.91/ 10)

=5.3

The 98% confidence interval estimate of the population mean is,

- E < < + E

37.1 - 5.3 < <37.1 + 5.3

31.8 < < 42.4

31.8 , 42.4