The amounts (in millilitres) of juice in eight randomly selected juice bottles are: $443.55467.21443.55449.46470.16464.25443.55467.21\begin{array} { l l l l } 443.55 & 467.21 & 443.55 & 449.46 \\470.16 & 464.25 & 443.55 & 467.21\end{array}$ Construct a 98% confidence interval for the mean amount of juice in all such bottles.

A) (468.09,471.05)
B) (441.48,471.05)
C) (443.28,468.96)
D) (468.09,482.22)
E) (441.48,468.09)

Question

The amounts (in millilitres) of juice in eight randomly selected juice bottles are:

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}

Construct a 98% confidence interval for the mean amount of juice in all such bottles.

A) (468.09,471.05)
B) (441.48,471.05)
C) (443.28,468.96)
D) (468.09,482.22)
E) (441.48,468.09)

Kheyzel Benedicto · Accepted Answer

Final Answer : C, Explanation :  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 = 455.125s = √[(∑(x - x̄)²)/(n - 1)] = √[(475.875)/7] = 6.526The 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.

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