Question

A random sample of 11 fields of spring wheat has a mean yield of 20.2 bushels per acre and standard deviation of 5.19 bushels per acre. Determine the 99% confidence interval for the true mean yield. Assume the population is approximately normal

Answer 1

`CI=(15.2,25.1)`

Explanation:

From the Question We are told that

Sample size
`n=11`

Mean
`\=x =20.2`

Standard Deviation
`\sigma=5.19`

Generally the equation for Critical Value is mathematically given by

`Critical\ value = t_(\alpha/2,(n-1))`

`Critical\ Value=t_(0.01/2,10)`

`Critical\ Value=3.1693`

Generally he 99% confidence interval for the mean yield is

`CI=\bar{X}\pm t_(\alpha/2,(n-1))S/{√(n)}`

`CI=20.2\pm 3.1693*5.19/{√(11)}`

`CI=20.2\pm4.9595`

`CI=(15.2,25.1)`