
a.
Obtain the sample
Find the coefficient of variation for the lengths and widths.
a.

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

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,
Here, one has to find 95% confidence interval for μ.
The confidence level, c=0.95.
Using?Table?5: Areas of a Standard
From Part (a), the value of a sample standard deviation,
The margin of error is as follows:
The 95% confidence interval is obtained as shown below:
The 95% confidence interval for μ is (406.5, 470.3).
(c)
Obtain the number of more little neck clams that would be needed in a sample.
(c)

Answer to Problem 1DH
The additional little neck clams needed is 357.
Explanation of Solution
Calculation:
In this scenario, it is known that
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
The number of more little neck clams that would be needed in a sample is obtained as follows:
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
d.
Obtain the 95% confidence interval for the population mean width of all Garrison Bay little neck clams.
d.

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,
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,
The 95% confidence interval is obtained as follows:
The 95% confidence interval for μ is (354.03, 413.23).
(e)
Find the number of more little neck clams that would be needed in a sample.
(e)

Answer to Problem 1DH
The additional little neck clams needed is 272.
Explanation of Solution
Calculation:
Here, it is known that
The number of more little neck clams that would be needed in a sample is obtained as follows:
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
f)
State whether the sample measurements length and width are independent or dependent.
f)

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.
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Chapter 7 Solutions
Understandable Statistics: Concepts and Methods
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