The amounts (in millilitres)of juice in eight randomly selected juice bottles are: $$\begin{array} { l l l l }
443.55 & 467.21 & 443.55 & 449.46 \
470.16 & 464.25 & 443.55 & 467.21
\end{array}$$ 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: $$\begin{array} { l l l l } 
443.55 & 467.21 & 443.55 & 449.46 \
470.16 & 464.25 & 443.55 & 467.21
\end{array}$$ Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Quizplus · Accepted Answer

We can use the formula: 
CI = x̄ ± tα/2(s/√n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and tα/2 is the t-value with α/2 = 0.01 and degrees of freedom (df) = n - 1.
Calculating the sample mean and sample standard deviation:
x̄ = (455 + 459 + 446 + 463 + 452 + 449 + 460 + 467)/8 = 455.125
s = √[(∑(x - x̄)²)/(n - 1)] = √[(475.875)/7] = 6.526
The t-value with α/2 = 0.01 and df = 7 is 3.499.
Substituting into the formula:
CI = 455.125 ± 3.499(6.526/√8) ≈ (443.28, 468.96)
Therefore, choice C is the best answer.

The Amounts (In Millilitres)of Juice in Eight Randomly Selected Juice 443.55467.21443.55449.46470.16464.25443.55467.21\begin{array} { l l l l } 443.55 & 467.21 & 443.55 & 449.46 \\470.16 & 464.25 & 443.55 & 467.21\end{array}443.55470.16​467.21464.25​443.55443.55​449.46467.21​

The Amounts (In Millilitres)of Juice in Eight Randomly Selected Juice $\begin{array} { l l l l } 443.55 & 467.21 & 443.55 & 449.46 \\470.16 & 464.25 & 443.55 & 467.21\end{array}$