Question

A random sample of 28 fields of spring wheat has a mean yield of 44.7 bushels per acre and standard deviation of 6.96 bushels per acre. Determine the 95% 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 95%95% confidence interval. Round your answer to one decimal place.

LOWER END POINT: ___________ UPPER END POINT:_______________

Answer 1

Solution :

Given that,

= 44.7

s = 6.96

n = 28

Degrees of freedom = df = n - 1 = 28 - 1 = 27

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,27 = 2.052

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

= 2.052 * (6.96 / 28)

= 2.7

The 95% confidence interval estimate of the population mean is,

- E < < + E

44.7 - 2.7 < < 44.7 + 2.7

42.0 < < 47.4

Lower end point = 42.0

Upper end point = 47.4