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A random sample of 11 fields of rye has a mean yield of 20.1 bushels per acre and standard deviation of 7.66 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.

Respuesta :

abidemiokin

Answer:

17.1≤x≤23.1

Step-by-step explanation:

The formula for calculating the confidence interval is expressed as;

CI = x ± z*s/√n

x is the mean yield = 20.1

z is the 80% z-score = 1.282

s is the standard deviation = 7.66

n is the sample size =  11

Substitute

CI = 20.1 ± 1.282*7.66/√11

CI = 20.1 ± 1.282*7.66/3.3166

CI = 20.1 ± 1.282*2.3095

CI =  20.1 ±2.9609

CI = (20.1-2.9609, 20.1+2.9609)

CI = (17.139, 23.0609)

hence the required confidence interval to 1dp is 17.1≤x≤23.1

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