Question

A random sample of 7 fields of corn has a mean yield of 31.0 bushels per acre and standard deviation of 7.05 bushels per acre. Determine the 90% 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

Solution :

Given that,

= 31.0

s = 7.05

n = 7

Degrees of freedom = df = n - 1 = 7 - 1 =6

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,6 =1.943

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

= 1.943 * (7.05 / 7)

= 5.177.

Margin of error = 5.177

The 90% confidence interval estimate of the population mean is,

- E < < + E

31 - 5.177 < < 31 + 5.177

25.823 < < 36.177

(25.823, 36.177)