The amounts (in millilitres)of juice in eight randomly selected juice bottles are:
Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Question

The amounts (in millilitres)of juice in eight randomly selected juice bottles are: 
Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Quizplus · Accepted Answer

Assuming a normal distribution, the point estimate of the mean amount of juice is the sample mean, which is 455.125 mL. 
The standard error can be calculated as: 
SE = s / sqrt(n) = 7.993 / sqrt(8) = 2.828 
Using a t-distribution with 7 degrees of freedom (df=n-1), the critical value for a 98% confidence level is 2.998. 
The margin of error can be calculated as: 
ME = critical value x SE = 2.998 x 2.828 = 8.486 
Therefore, the 98% confidence interval for the population mean is:
455.125 - 8.486 < μ < 455.125 + 8.486 
446.639 < μ < 463.611 
Rounded to two decimal places, the interval is (443.28, 468.96). Therefore, the best choice is C.

The Amounts (In Millilitres)of Juice in Eight Randomly Selected Juice