Chapter 8, Problem 1DHGP

Garrison Bay is a small hay in Washington stale. A popular recreational activity in the bay is clam digging. Forseveral years, this harvest has been monitored and the size distribution of clams recorded. Data for lengths and widths of little neck clams (Protothacastaminea) were recorded by a method of systematic sampling in a study done by S. Scherba and V. F. Gallucci ("The Application of Systematic Sampling to a Study of Infaunal Variation in a Soft Substrate Intertidal Environment." Fishery Bulletin, Vol. 74, pp 937-948). The data in Tables 8-4 and 8-5 are the lengths and widths for 35 little neck clams.

(a) Use a calculator to compute the sample mean and sample standard deviation for the lengths and widths Compute the coefficient of variation for each.

(b) Compute a 95% confidence interval for the population mean length of all Garrison Bay little neck clams.

(c) How many more little neck clams would be needed in a sample if you wanted to be 95% sure that the sample mean length is within a maximal margin of error of 10 mm of the population mean length?

TABLE 8-4 Lengths of Little Neck Clams (mm)

530	517	505	512	487	481	485	479	452	468
459	449	472	471	455	394	475	335	508	486
474	465	420	402	410	393	389	330	305	169
91	537	519	509	511

TABLE 8-5 Widths of Little Neck Clams (mm)

494	477	471	413	407	427	408	430	395	417
394	397	402	401	385	338	422	288	464	436
414	402	383	340	349	333	356	268	264	141
11	498	456	433	447

(d) Compute a 95% confidence interval for the population mean width of all Garrison Bay little neck clams.

(c) How many more little neck claim would he needed in a sample if you wanted to be 95% sure that the sample mean width is within a maximal margin of error of 10 mm of the population mean width?

(0 Thesame 35 clams mere used for measures of length and width. Are the sample measurements length and width independent or dependent? Why?

To determine

To find: Thesample mean, sample standard deviation and the coefficient of variation for the lengths and widths.

Answer to Problem 1DHGP

Solution:

Variable	Mean	Standard deviation	Coefficient of variation
length_(mm)	438.4	96.4	21.98
width_(mm)	383.6	89.4	23.3

Explanation of Solution

Calculation:

Let x represents the lengths of little neck clams and y represent the widths of little neck clams.

To find sample mean, sample standard deviation and the coefficient of variationfor x, yby using Minitab, follow the instructions:

Step 1: Enter the data in Minitab Software.

Step 2: Go to Stat > Basic statistics > Display Descriptive Statistics

Step 3: Select x, y in variable.

Step 4: Go to ‘statistics’ choose Mean, Standard deviation, Coefficient of variation.

The result is obtained below:

Variable	Mean	Standard deviation	Coefficient of variation
length_(mm)	438.4	96.4	21.98
width_(mm)	383.6	89.4	23.3

To determine

To find: The95% confidence interval for the population mean length of all Garrison Bay little neck clams.

Answer to Problem 1DHGP

Solution:

The95% confidence interval for the population mean length of all Garrison Bay little neck clams is between 406.5 and 470.3.

Explanation of Solution

Calculation:

Let x represents the lengths of little neck clams, $\overline{x}$ be the sample mean length of all Garrison Bay little neck clams, $\overline{x}=438.4$ and sample size $n=35$. Assume that $\sigma$ is unknown, so we use s be the estimate of $\sigma$.

We have to find 95% confidence interval for $\mu$,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.96\\ s_x=96.4\\ n=35\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.96\frac{96.4}{\sqrt{35}}\\ E=31.937\\ E\approx 31.9\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 438.4-31.9<\mu <438.4+31.4\\ 406.5<\mu <470.3\end{array}$

The 95% confidence interval for $\mu$ is (406.5, 470.3).

To determine

To find: The number of more little neck clams would be needed in a sample.

Answer to Problem 1DHGP

Solution:

The additional little neck clams needed is357.

Explanation of Solution

Calculation:

We have, $s_x=96.4$, $n=35$, E = 10.

We have to find the sample size n,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=196\\ s=96.4\\ E=10\\ n={\left(\frac{z_{c}s}{E}\right)}^2\\ n={\left(\frac{1.96(96.4)}{10}\right)}^2\\ n=356.9984\\ n\approx 357\end{array}$

The sample size needed to be 95% sure that the sample mean is within a maximal margin of error of 10 mm of the population mean length is 357.

So, the additional little neck clams needed $=357-35=322$

To determine

To find: The95% confidence interval for the population mean width of all Garrison Bay little neck clams.

Answer to Problem 1DHGP

Solution:

The95% confidence interval for the population mean width of all Garrison Bay little neck clams is between 354.03 and 413.23.

Explanation of Solution

Calculation:

Let y represents the widths of little neck clams, $\overline{y}$ be the sample meanwidth of all Garrison Bay little neck clams, $\overline{y}=383.6$ and sample size $n=35$. Assume that $\sigma$ is unknown, so we use s be the estimate of $\sigma$.

We have to find 95% confidence interval for $\mu$,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=1.96\\ s_y=89.4\\ n=35\\ E=z_c\frac{s}{\sqrt{n}}\\ E=1.96\frac{89.4}{\sqrt{35}}\\ E=29.618\\ E\approx 29.6\end{array}$

95% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 383.6-29.6<\mu <383.6+29.6\\ 354.03<\mu <413.23\end{array}$

The 95% confidence interval for $\mu$ is (354.03, 413.23).

To determine

To find: The number of more little neck clams would be needed in a sample.

Answer to Problem 1DHGP

Solution:

The additional little neck clams needed is 272.

Explanation of Solution

Calculation:

We have, $s_y=89.6$, $n=35$, E = 10.

We have to find the sample size n,

$\begin{array}{l}c=0.95\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.95}=196\\ s_y=89.4\\ E=10\\ n={\left(\frac{z_{c}s}{E}\right)}^2\\ n={\left(\frac{1.96(89.4)}{10}\right)}^2\\ n=307.0\\ n\approx 307\end{array}$

The sample size needed to be 95% sure that the sample mean is within a maximal margin of error of 10 mm of the population mean length is 307.

So, the additional little neck clams needed $=307-35=272$

To determine

To explain: Whether the sample measurements length and width are independent or dependent.

Answer to Problem 1DHGP

Solution: The sample measurements length and width are dependent.

Explanation of Solution

The same 35 clams were used for measures of length and width. The sample measurements length and width are dependent because as the 95% confidence interval for length increases at same level of confidence, the width is decreased.