Question

A random sample of 25 fields of rye has a mean yield of 31.2 bushels per acre and standard deviation of 6.99 bushels per acre. Determine the 80% 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.

Answer 1

Critical value is 1.318 and the confidence interval is 29.357 to 33.042

Explanation:

n = 25

x = 31.2

s = 6.99

Degree of freedom = n-1= 25-1 =24

Confidence level = 0.8

So, α = 1- 0.8= 0.2

t critical =
`t_{(\alpha)/(2),df}=t_{(0.2)/(2),24}=t_(0.10,24)=1.318`

Formula of confidence interval :
`\bar{x}-t_{(\alpha)/(2),df} * (s)/(√(n))` to
`\bar{x}+t_{(\alpha)/(2),df} * (s)/(√(n))`

Confidence interval :
`31.2-1.318 * (6.99)/(√(25))` to
`31.2+1.318 * (6.99)/(√(25))`

Confidence interval :
`29.357` to
`33.042`

Hence critical value is 1.318 and the confidence interval is 29.357 to 33.042