Question

Ethanol Railroad Tariffs. An ethanol railroad tariff is a fee charged for shipments of ethanol on public railroads. The Agricultural Marketing Service publishes tariff rates for railroad-car shipments of ethanol in the Biofuel Transportation Database. Assuming that the standard deviation of such tariff rates is $\backslash (1,150$, determine the probability that the mean tariff rate of $500$randomly selected railroad car shipments of ethanol will be within $\backslash )100$of the mean tariff rate of all railroad-car shipments of ethanol. Interpret your answer in terms of sampling error.

Short answer

There is a $94$percent chance that the sampling error in predicting the mean tariff rate of all railroad-car ethanol shipments by a sample of$500$ railroad-car ethanol shipments is within $\$100$

Step-by-step solution

Given information

The population mean tariff rate for all ethanol railroad-car shipments, and $\left(\sigma \right)$be the population standard deviation is $\$1,150$In addition, the sample size $n$is $500$, and the sample mean tariff rate $\left(\overline{x}\right)$is regularly distributed.

Calculation

Let $\left(\sigma \right)$be the population standard deviation for all ethanol railroad-car shipments, and $\mu$be the population mean tariff rate. Furthermore, the sample size is $\left(n\right)$and the sample means tariff rate is evenly distributed.

The average tariff rate in the sample ${\mu }_x=\mu$and sample standard deviation as,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{1,150}{\sqrt{500}} \\ =\frac{1,150}{22.3607} \\ =51.43 \end{gathered}$$

Calculation

Calculate the likelihood that the mean tariff rate of $500$randomly selected ethanol railroad-car shipments is within $\$100$of the mean tariff rate of all ethanol railroad-car shipments.

The z-score for $\mu -100$is,

$$\begin{gathered} z=\frac{\mu -100-\mu }{51.43} \\ =\frac{-100}{51.43} \\ =-1.94 \end{gathered}$$

The z-score for $\mu +100$is

$$\begin{gathered} z=\frac{{\mu }_{-}100-\mu }{51.43} \\ =\frac{100}{51.43} \\ =1.94 \end{gathered}$$

Calculation

To find the area between the z-scores, use Table II: Areas under the standard normal curve.

$1.94$is the area to the left of the $z-$score $0.0262$

$1.94$is the area to the left of the $z-$score $0.9738$

The area between $z-$scores $=(Areatotheleftof1.94)-(Areatotheleftof-1.94)$

$$\begin{gathered} =0.9738-0.0262 \\ =0.9476 \end{gathered}$$

As a result, the probability that the mean tariff rate of randomly selected ethanol railroad-car shipments is within $\$100$the mean tariff rate of all ethanol railroad-car shipments is $0.9476$