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.
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 = t0.10,8 = 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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