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Scallops, sampling, and the law. Interfaces (March鈥揂pril 1995) presented the case of a ship that fishes for scallops off the coast of New England. In order to protect baby scallops from being harvested, the U.S. Fisheries and Wildlife Service requires that 鈥渢he average meat per scallop weigh at least 136 of a pound.鈥� The ship was accused of violating this weight standard. Author Arnold Barnett lays out the scenario:

The vessel arrived at a Massachusetts port with 11,000 bags of scallops, from which the harbormaster randomly selected 18 bags for weighing. From each such bag, his agents took a large scoopful of scallops; then, to estimate the bag鈥檚 average meat per scallop, they divided the total weight of meat in the scoopful by the number of scallops it contained. Based on the 18 [numbers] thus generated, the harbormaster estimated that each of the ship鈥檚 scallops possessed an average of 139 of a pound of meat (that is, they were about seven percent lighter than the minimum requirement). Viewing this outcome as conclusive evidence that the weight standard had been violated, federal authorities at once confiscated 95 percent of the catch (which they then sold at auction). The fishing voyage was thus transformed into a financial catastrophe for its participants. The actual scallop weight measurements for each of the 18 sampled bags are listed in the table below. For ease of exposition, Barnett expressed each number as a multiple of of a pound, the minimum permissible average weight per scallop. Consequently, numbers below 1 indicate individual bags that do not meet the standard. The ship鈥檚 owner filed a lawsuit against the federal government, declaring that his vessel had fully complied with the weight standard. A Boston law firm was hired to represent the owner in legal proceedings, and Barnett was retained by the firm to provide statistical litigation support and, if necessary, expert witness testimony.

0.93	0.88	0.85	0.91	0.91	0.84	0.90	0.98	0.88
0.89	0.98	0.87	0.91	0.92	0.99	1.14	1.06	0.93

1. Recall that the harbormaster sampled only 18 of the ship鈥檚 11,000 bags of scallops. One of the questions the lawyers asked Barnett was, 鈥淐an a reliable estimate of the mean weight of all the scallops be obtained from a sample of size 18?鈥� Give your opinion on this issue.
2. As stated in the article, the government鈥檚 decision rule is to confiscate a catch if the sample mean weight of the scallops is less than 136 of a pound. Do you see any flaws in this rule?
3. Develop your own procedure for determining whether a ship is in violation of the minimum-weight restriction. Apply your rule to the data. Draw a conclusion about the ship in question.

Step by step solution

01

Given information

In Interfaces, the cases involved a ship that fishes for scallops off the coast of New England. In order to protect baby scallops from being harvested, the fisheries require that 鈥渢he average meat per scallop weight at least$\frac{1}{36}$ of a pound.鈥�

02

Finding the estimate

a.

It is given that harbor master sampled only 18 of the ship鈥檚 11,000 bags of scallops.

The proportion of scallops in the sample will be

$\begin{array}{rcl}\mathrm{p}& =& \frac{18}{11000}\\ & =& 0.00164\end{array}$

Since, the proportion of sampled scallops is very small (almost equivalent to zero). It may conclude that the sampled scallops do not reasonably represent the whole population.

03

Determining flaws in a rule

b.

It is stated in the article that the government鈥檚 decision rule is to confiscate a catch if the sample mean weight of the scallops is less than$\frac{1}{36}$ of a pound.

Here in this case, the measure of reliability is missing.

04

Applying the rule and concluding

c.

The actual scallop weights measurements for each of the 18 sampled bags are given by expressing each number as a multiple of $\frac{1}{36}$of a pound and are given as follows:

0.93	0.88	0.85	0.91	0.91	0.84	0.90	0.98	0.88
0.89	0.98	0.87	0.91	0.92	0.99	1.14	1.06	0.93

The 95% confidence interval for the mean weight of scallops is defined as follows:

$\stackrel{\mathbf{炉}}{\mathbf{x}}\mathbf{卤}{\mathbf{t}}_{\mathbf{伪}\overline{)\mathbf{2}\mathbf{,}\left(n-1\right)}}\mathbf{脳}\frac{\mathbf{s}}{\sqrt{\mathbf{n}}}$

Also, the sample means,

$\stackrel{\mathbf{炉}}{\mathbf{x}}\mathbf{=}\frac{{\displaystyle \underset{\mathbf{i}\mathbf{=}\mathbf{1}}{\overset{\mathbf{n}}{\mathbf{鈭�}{\mathbf{x}}_{\mathbf{i}}}}}}{\mathbf{n}}$

Therefore,

$\begin{array}{rcl}\stackrel{炉}{\mathrm{x}}& =& \frac{{\displaystyle \underset{\mathrm{i}=1}{\overset{\mathrm{n}}{鈭�{\mathrm{x}}_{\mathrm{i}}}}}}{18}\\ & =& \frac{16.7652}{18}\\ & =& 0.9314\end{array}$

The sample standard deviation,

role="math" localid="1658315856500" $\mathbf{s}\mathbf{=}\sqrt{\frac{{\displaystyle \underset{i=1}{\overset{n}{鈭�}}}{\left(x_i-\stackrel{炉}{x}\right)}^2}{\mathbf{n}\mathbf{-}\mathbf{1}}}$

Therefore,

$$\begin{gathered} \mathrm{s}=\sqrt{\frac{{\displaystyle \underset{\mathrm{i}=1}{\overset{18}{鈭�}}{\left({\mathrm{x}}_{\mathrm{i}}-\stackrel{炉}{\mathrm{x}}\right)}^2}}{18-1}} \\ \mathrm{s}=\sqrt{\frac{0.09680}{17}} \\ \mathrm{s}=0.07546 \end{gathered}$$

The critical value of t at $\left(\mathrm{n}-1\right)=\left(18-1\right)$degrees of freedom, ${\mathrm{t}}_{0.05\overline{)2.17}}=2.11$

Then, the 95% confidence interval is given by,

$$\begin{gathered} 0.9314卤2.110脳\frac{0.07546}{\sqrt{18}} \\ 0.9314卤0.0272 \\ \left(0.9042,0.9586\right) \end{gathered}$$

In terms of original weights, the 95% confidence interval is given by $\left(0.02511,0.02662\right)$.

Because, the 95% confidence interval does not include$\frac{1}{36}$ of a pound and is less than the specified value, it can say that the ship is not fulfilled the government鈥檚 rule.

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