Question

A random sample of 9 fields of corn has a mean yield of 35.8 bushels per acre and standard deviation of 2.79 bushels per acre. Determine the 80% confidence interval for the true mean yield. Assume the population is approximately normal. Round your answer to one decimal place.

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 35.8

sample standard deviation = s = 2.79

sample size = n = 9

Degrees of freedom = df = n - 1 = 8

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t_{
/2,df} = t_{0.10,8} = 1.397

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

= 1.397* (2.79 / 9)

= 1.3

The 80% confidence interval estimate of the population mean is,

- E < < + E

35.8 - 1.3< < 35.8 + 1.3

34.5 < < 37.1

**The 80% confidence interval for the true mean yield is
(34.5 , 37.1)**

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