Lobster trap placement. An observational study of teams fishing for the red spiny lobster in Baja California Sur, Mexico, was conducted and the results published in Bulletin of Marine Science(April 2010). One of the variables of interest was the average distance separating traps—called trap spacing—deployed by the same team of fishermen. Trap-spacing measurements (in meters) for a sample of seven teams of red spiny lobster fishermen are shown in the accompanying table. Of interest is the mean

trap spacing for the population of red spiny lobster fishermen fishing in Baja California Sur, Mexico.

93

99

105

94

82

70

86

1. Identify the target parameter for this study.
2. Compute a point estimate of the target parameter.
3. What is the problem with using the normal (z) statistic to find a confidence interval for the target parameter?
4. Find a 95% confidence interval for the target parameter.
5. Give a practical interpretation of the interval, part d.
6. What conditions must be satisfied for the interval, part d, to be valid?

There is a study about fishing for red spiny lobster. A variable of interest is the average distance of the separate traps for fishing. Trap spacing measurements of the sample of seven teams are,

a.

The sample size is\(n = 7\).

The target parameter for the given study of fishing red spiny lobster is the mean trap spacing for the population.

b.

Let’s consider the point estimator of the target variable is\(\hat p\).

So,

\(\begin{aligned}\hat p &= \frac{{93 + 99 + 105 + 94 + 82 + 70 + 86}}{7}\\ &= 89.85\end{aligned}\)

Thus, the point estimate of the target variable is 89.85.

c.

As there is a sample of size 7 with their mean. So, the sample size is very small (less than 15), and also there is no considered population variance. So, there cannot be used the normal(z) statistic to find the confidence interval for the target parameter.

d.

The confidence interval is 95%.

Therefore, the level of significance\(\alpha = 1 - 0.95 = 0.05\).

So, from the t-table, the statistic\({t_{\alpha ,\left( {n - 1} \right)}} = {t_{0.05,6}} = 2.36\).

The sample mean is the point estimator. So,\(\hat p = \bar x\)

Now the standard deviation of the sample is,

\(\begin{aligned}s &= \sqrt {\frac{{\sum\limits_{i = 1}^7 {\left( {{x_i} - \bar x} \right)} }}{{n - 1}}} \\ &= \sqrt {\frac{{810.85}}{6}} \\ &= 11.62 \end{aligned}\)

Therefore, the confidence interval is\(\left( {\bar x \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{s}{{\sqrt n }}} \right)\).

Thus,

\(\begin{aligned}\left( {\bar x \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{s}{{\sqrt n }}} \right) &= \left( {89.85 \pm {t_{0.05,6}} \times \frac{{11.62}}{{\sqrt 7 }}} \right)\\ &= \left( {\left( {89.85 - 2.36 \times 4.39} \right),\left( {89.85 + 2.36 \times 4.39} \right)} \right)\\ &= \left( {79.48,100.21} \right) \end{aligned}\)

Therefore, the 95% confidence interval is (79.48,100.21).

e.

The population mean\(\bar x = 89.85\). This is between the 95% confidence interval. So, there can be concluded that the population means will be within this interval.

But the information, given in the study is random and there is a pattern to the study’s behavior in the possible samples. Each sample is giving a different sample proportion and interval. But there is the confidence that the margin of error will satisfy 95% of all samples.

f.

The condition that must be satisfied for the interval in part ‘d’, to be valid is the assumption that the population follows the normal distribution, so, the samples come from the normal distribution.