Chapter 7, Problem 1DH

a.

To determine

Obtain the sample mean and sample standard deviation for the lengths and widths.

Find the coefficient of variation for the lengths and widths.

Answer to Problem 1DH

The sample mean and sample standard deviation for the lengths and widths are given below:

Variable	Mean	Standard deviation	Coefficient of variation
Length	438.4	96.4	21.98
Width	383.6	89.4	23.3

Explanation of Solution

Calculation:

Let x represent 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 variation for x, y  using MINITAB.

Step-by-step procedure to obtain the descriptive measures using MINITAB:

• Enter the data in MINITAB Software.
• Select Stat > Basic statistics > Display Descriptive Statistics
• Select x and y in variable.
• In ‘statistics’ choose Mean, Standard deviation, Coefficient of variation.

Output obtained using MINITAB is given below:

(b)

To determine

Obtain the 95% confidence interval for the population mean length of all Garrison Bay little neck clams.

Answer to Problem 1DH

The 95% 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 represent 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; thus, use s as the estimate of $\sigma$.

Here, one has to find 95% confidence interval for μ.

The confidence level, c=0.95.

Using?Table?5: Areas of a Standard Normal Distribution?of?Appendix II, the critical value of 95% confidence level is $z_{0.95}=1.96$.

From  Part (a), the value of a sample standard deviation, $s_{\overline{x}}=96.4$.

The margin of error is as follows:

$\begin{array}{c}E=z_c\frac{s}{\sqrt{n}}\\ =z_{0.95}\frac{s}{\sqrt{n}}\\ =1.96\frac{96.4}{35}\\ =31.937\\ \approx 31.9\end{array}$

The 95% confidence interval is obtained as shown below:

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

The 95% confidence interval for μ is (406.5, 470.3).

(c)

To determine

Obtain the number of more little neck clams that would be needed in a sample.

Answer to Problem 1DH

The additional little neck clams needed is 357.

Explanation of Solution

Calculation:

In this scenario, it is known that $s_{\overline{x}}=96.4, n=35, E =10$.

One has to find the sample size n.

The confidence level, c=0.95.

Using?Table?5: Areas of a Standard Normal Distribution?of?Appendix II, the critical value of 95% confidence level is $z_{0.95}=1.96$.

The number of more little neck clams that would be needed in a sample is obtained as follows:

$\begin{array}{c}n={\left(\frac{z_{c}s}{E}\right)}^2\\ =\left(\frac{z_{0.95}s}{E}\right)\\ n={\left(\frac{1.96\left(96.4\right)}{10}\right)}^2\\ n=356.9984\\ n\approx 357\end{array}$

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.

Thus, the additional little neck clams needed is $\underset{\_}{322\left(=357-35\right)}$.

d.

To determine

Obtain the 95% confidence interval for the population mean width of all Garrison Bay little neck clams.

Answer to Problem 1DH

The 95% 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 represent the width of little neck clams, $\overline{y}$ be the sample mean width of all Garrison Bay little neck clams, $\overline{y}=383.6$ and sample size $n=35$. Assume that $\sigma$ is unknown; thus, one uses s as the estimate of $\sigma$.

Here, one has to find 95% confidence interval for μ.

The confidence level, c=0.95.

Using?Table?5: Areas of a Standard Normal Distribution?of?Appendix II, the critical value of 95% confidence level is, $z_{0.95}=1.96$.

$\begin{array}{c}E=z_c\frac{s}{\sqrt{n}}\\ E=\frac{1.96\times 89.4}{\sqrt{35}}\\ E=29.618\\ E\approx 29.6\end{array}$

The 95% confidence interval is obtained as follows:

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

The 95% confidence interval for μ is (354.03, 413.23).

(e)

To determine

Find the number of more little neck clams that would be needed in a sample.

Answer to Problem 1DH

The additional little neck clams needed is 272.

Explanation of Solution

Calculation:

Here, it is known that $s_{\overline{y}}=89.6, n=35, E =10$.

The number of more little neck clams that would be needed in a sample is obtained as follows:

$\begin{array}{c}n={\left(\frac{z_{c}s}{E}\right)}^2\\ n={\left(\frac{1.96\left(89.4\right)}{10}\right)}^2\\ n=307.0\\ n\approx 307\end{array}$

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.

Thus, the additional little neck clams needed is $272\left(=307-35\right)$.

f)

To determine

State whether the sample measurements length and width are independent or dependent.

Answer to Problem 1DH

The sample measurements length and width are dependent.

Explanation of Solution

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