Question

The American Sugar Producers Association wants to estimate

Answer 1

Answer: 39.308 pounds

Step-by-step explanation:

We assume that the given population is normally distributed.

Given : Significance level : $\alpha: 1-0.98=0.02$

Sample size : n= 12, which is  small sample (n<30), so we use t-test.

Critical value (by using the t-value table)=$t_{n-1, \alpha/2}=t_{11,0.01}=2.718$

Sample mean : $\overline{x}=50$  

Standard deviation : $\sigma= 20$

The lower bound of confidence interval is given by :-

$\overline{x}-t_{(n-1,\alpha/2)}\dfrac{\sigma}{\sqrt{n}}$

i.e. $55-(2.718)\dfrac{20}{\sqrt{12}}$

$=55-15.6923803166\approx55-15.692=39.308$

Hence, the lower bound for the 98%  confidence interval for the mean yearly sugar consumption= 39.308 pounds