Answer: 39.308 pounds
Step-by-step explanation:
We assume that the given population is normally distributed.
Given : Significance level : [tex]\alpha: 1-0.98=0.02[/tex]
Sample size : n= 12, which is small sample (n<30), so we use t-test.
Critical value (by using the t-value table)=[tex]t_{n-1, \alpha/2}=t_{11,0.01}=2.718[/tex]
Sample mean : [tex]\overline{x}=50[/tex]
Standard deviation : [tex]\sigma= 20[/tex]
The lower bound of confidence interval is given by :-
[tex]\overline{x}-t_{(n-1,\alpha/2)}\dfrac{\sigma}{\sqrt{n}}[/tex]
i.e. [tex]55-(2.718)\dfrac{20}{\sqrt{12}}[/tex]
[tex]=55-15.6923803166\approx55-15.692=39.308[/tex]
Hence, the lower bound for the 98% confidence interval for the mean yearly sugar consumption= 39.308 pounds
