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A random sample of 77 fields of durum wheat has a mean yield of 27.427.4 bushels per acre and standard deviation of 5.755.75 bushels per acre. Determine the 99%99% 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

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Answer:

[ 21.79, 33.0 ]

Step-by-step explanation:

Given:

Sample size, n = 7

Sample mean, μ = 27.4

Standard deviation, σ = 5.75

Confidence level is 99%

also, population is normal i.e normally distributed

Thus,

Confidence interval = μ ± [tex]z\frac{\sigma}{\sqrt n}[/tex]

For confidence level of 99%

z-value = 2.58

Therefore,

Lower limit of Confidence interval = μ - [tex]2.58\times\frac{5.75}{\sqrt{7}}[/tex]

or

Lower limit of Confidence interval = 27.4 - [tex]2.58\times\frac{5.75}{\sqrt{7}}[/tex]

or

Lower limit of Confidence interval = 21.79

Upper limit of Confidence interval = μ + [tex]2.58\times\frac{5.75}{\sqrt{7}}[/tex]

or

Upper limit of Confidence interval = 27.4 + [tex]2.58\times\frac{5.75}{\sqrt{7}}[/tex]

or

Upper limit of Confidence interval = 33.00

Hence, confidence interval = [ 21.79, 33.0 ]

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