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

Answer:

$CI=(15.2,25.1)$

Step-by-step 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

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

$CI=20.2\pm 3.1693*5.19/{\sqrt{11}}$

$CI=20.2\pm4.9595$

$CI=(15.2,25.1)$