Chapter 14, Problem 14.1.4RE

To determine

To divide: The 28 storms into 4 subgroups.

To select: A random sample of 3 storms from each group.

To compute: The means for maximum wind speeds.

To compare: The means to the population mean.

Explanation of Solution

Given info:

In the 2005 Atlantic hurricane season the maximum wind speed and classifications for different storms are given in the data.

Calculation:

First arrange the data according to the serial number,

Serial No.	Name	Max. wind	Classification
1	Arlene	70	Storm
2	Bret	40	S
3	Cindy	75	Hurricane
4	Dennis	150	H
5	Emily	160	H
6	Franklin	70	S
7	Gert	45	S
8	Harvey	65	S
9	Irene	105	H
10	Jose	50	S
11	Katrina	175	H
12	Lee	40	S
13	Maria	115	H
14	Nate	90	H
15	Ophelia	85	H
16	Philippe	70	H
17	Rita	175	H
18	Stan	80	H
19	Unnamed	50	S
20	Tammy	50	S
21	Vince	75	H
22	Wilma	175	H
23	Alpha	50	S
24	Beta	115	H
25	Gamma	55	S
26	Delta	70	S
27	Epsilon	85	H
28	Zeta	65	S

$\text{S}=\text{Storm, H}=\text{Hurricane}$

Procedure for dividing the 28 storms into 4 subgroups:

Step1:

Divide the population into two groups. From the 28 storms, divide the observations according to classification- Storm and Hurricane. There are 15 Hurricanes and 13 Storms:

Group 1 (H)		Group 2 (S)
Name	Max. wind	Name	Max. wind
Cindy	75	Arlene	70
Dennis	150	Bret	40
Emily	160	Franklin	70
Irene	105	Gert	45
Katrina	175	Harvey	65
Maria	115	Jose	50
Nate	90	Lee	40
Ophelia	85	Unnamed	50
Philippe	70	Tammy	50
Rita	175	Alpha	50
Stan	80	Gamma	55
Vince	75	Delta	70
Wilma	175	Zeta	65
Beta	115
Epsilon	85

Step2:

Divide each group into two subgroups, based on the median value of max. wind.

For Group 1 (H), the median value of max. wind is 105. Thus, the 1^st subgroup will contain all storms with max. wind less than or equal to 105 and the 2^nd subgroup will contain all storms with max. wind greater than 105.

For Group 2 (S), the median value of max. wind is 50. Thus, the 3^rd subgroup will contain all storms with max. wind less than or equal to 50 and the 4^th subgroup will contain all storms with max. wind greater than 50.

The subgroups are:

SNO	Subgroup1	Subgroup2	Subgroup3	Subgroup4
1	Cindy	Dennis	Bret	Arlene
2	Irene	Emily	Gert	Franklin
3	Nate	Katrina	Jose	Harvey
4	Ophelia	Maria	Lee	Gamma
5	Philippe	Rita	Unnamed	Delta
6	Stan	Wilma	Tammy	Zeta
7	Vince	Beta	Alpha
8	Epsilon

From the s select a random sample of 3 storms by using random numbers.

Procedure for selecting 3 random samples from subgroup 1:

• From the figure 14.1(Table of random numbers), select a starting point.
• Here, the selected starting point is ‘7’ which is present in 1^st  row and 1^st  column. There are 8 names of storms in subgroup 1. Hence, select the random numbers in the range of ‘1’ to ‘8. Avoid repetition of numbers and 0.
• The random numbers starting from ‘7’ are:

  7, 2, 1.

Thus, the random numbers are and corresponding names of the wind are,

Serial No.	Name
7	Vince
2	Irene
1	Cindy

The average maximum wind of above selected sample is calculated by using the following formula.

$\begin{array}{c}\text{Sample mean of Max}.\text{wind speed} =\frac{\text{Sum}\text{of}\text{all}\text{sample}\text{values}\text{of}\text{Max}\text{.wind}}{\text{Number}\text{of}\text{samples}}\\ =\frac{75+105+75}{3}\\ =\frac{255}{3}\\ =85.\\ \text{for sample from subgroup 1}\end{array}$

Thus, the sample mean for subgroup 1 is 85.

Procedure for selecting 3 random samples from subgroup 2:

• From the figure 14.1(Table of random numbers), select a starting point.
• Here, the selected starting point is ‘4’ which is present in 1^st row and 2^nd column. There are 7 names of storms in subgroup 2. Hence, select the random numbers in the range of ‘1’ to ‘7’. Avoid repetition of numbers and 0.
• The random numbers starting from ‘4’ are:

  4, 5, 1.

Thus, the random numbers are and corresponding names of the wind are,

Serial No.	Name
4	Maria
5	Rita
1	Dennis

The average maximum wind of above selected sample is calculated by using the following formula.

$\begin{array}{c}\text{Sample mean of Max}.\text{wind speed} =\frac{\text{Sum}\text{of}\text{all}\text{sample}\text{values}\text{of}\text{Max}\text{.wind}}{\text{Number}\text{of}\text{samples}}\\ =\frac{115+175+150}{3}\\ =\frac{440}{3}\\ =\mathrm{146.67.}\\ \text{for sample from subgroup 2}\end{array}$

Thus, the sample mean for subgroup 2 is 146.67.

Procedure for selecting 3 random samples from subgroup 3:

• From the figure 14.1(Table of random numbers), select a starting point.
• Here, the selected starting point is ‘7’ which is present in 5^th row and 13^th column. There are 7 names of storms in subgroup 3. Hence, select the random numbers in the range of ‘1’ to ‘7’. Avoid repetition of numbers and 0.
• The random numbers starting from ‘7’ are:

  7, 4, 2.

Thus, the random numbers are and corresponding names of the wind are,

Serial No.	Name
4	Alpha
5	Lee
1	Gret

The average maximum wind of above selected sample is calculated by using the following formula.

$\begin{array}{c}\text{Sample mean of Max}.\text{wind speed} =\frac{\text{Sum}\text{of}\text{all}\text{sample}\text{values}\text{of}\text{Max}\text{.wind}}{\text{Number}\text{of}\text{samples}}\\ =\frac{50+40+45}{3}\\ =\frac{135}{3}\\ =45.\\ \text{for sample from subgroup 3}\end{array}$

Thus, the sample mean for subgroup 3 is 45.

Procedure for selecting 3 random samples from subgroup 4:

• From the figure 14.1(Table of random numbers), select a starting point.
• Here, the selected starting point is ‘5’ which is present in 6^th row and 9^th column. There are 6 names of storms in subgroup 4. Hence, select the random numbers in the range of ‘1’ to ‘6’. Avoid repetition of numbers and 0.
• The random numbers starting from ‘5’ are:

  5, 2, 6.

Thus, the random numbers are and corresponding names of the wind are,

Serial No.	Name
5	Delta
2	Franklin
6	Zeta

The average maximum wind of above selected sample is calculated by using the following formula.

$\begin{array}{c}\text{Sample mean of Max}.\text{wind speed} =\frac{\text{Sum}\text{of}\text{all}\text{sample}\text{values}\text{of}\text{Max}\text{.wind}}{\text{Number}\text{of}\text{samples}}\\ =\frac{70+70+65}{3}\\ =\frac{205}{3}\\ =\mathrm{68.3.}\\ \text{for sample from subgroup 4}\end{array}$

Thus, the sample mean for subgroup 4 is 68.3.

The population mean of maximum wind is,

$\begin{array}{c}\text{Population mean} =\frac{\text{Sum}\text{of}\text{all}\text{values}\text{of}\text{Max}\text{.wind}}{\text{Number}\text{of}\text{obervations}}\\ =\frac{70+40+75+150+\mathrm{...}+85+65}{28}\\ =\frac{2,450}{28}\\ =\mathrm{87.5.}\end{array}$

Thus, the population mean is 87.5.

Comparison of means:

The sample mean for subgroup 1 is 85 and the population mean of maximum wind speed is 87.5.

The sample mean for subgroup 2 is 146.67 and the population mean of maximum wind speed is 87.5.

The sample mean for subgroup 3 is 45 and the population mean of maximum wind speed is 87.5.

The sample mean for subgroup 4 is 68.3 and the population mean of maximum wind speed is 87.5.

That implies that the sample means are smaller than the population mean for subgroup 3 an 4 whereas the sample mean is greater than the population mean for subgroup 2. But, for subgroup 1 sample mean is slightly smaller than population mean.

It is clear that there is some difference between the sample mean and the population mean.