Chapter 2, Problem 2.6AYK

(a)

To determine

To find: the five-number summary of the dataset.

Answer to Problem 2.6AYK

Statistics	Group Instruction	Individual instruction
Minimum:	78	128
First quartile:	117.50	159.75
Median:	147.50	191.50
Third quartile:	172.00	221.50
Maximum:	359	283

Explanation of Solution

Given:

Group Instruction	141	158	112	153	134	95	96	78	148
172	200	271	103	172	359	145	147	255
Individual instruction	128	195	188	158	227	198	163	164
159	128	283	226	223	221	220	160

Calculation:

The five-point statistics calculated using the Excel is given below:

Descriptive statistics
	Group Instruction	Individual instruction
count	18	16
mean	163.28	190.06
sample variance	4,992.21	1,755.66
sample standard deviation	70.66	41.90
minimum	78	128
maximum	359	283
range	281	155
1st quartile	117.50	159.75
median	147.50	191.50
3rd quartile	172.00	221.50
interquartile range	54.50	61.75
mode	172.00	128.00

The five-point summery is:

Statistics	Group Instruction	Individual instruction
Minimum:	78	128
First quartile:	117.50	159.75
Median:	147.50	191.50
Third quartile:	172.00	221.50
Maximum:	359	283

(b)

To determine

To construct: the box plot for the given datasets.

Explanation of Solution

Calculation:

  

  

The summery shown by the box plots is given below:

Statistics	Group Instruction	Individual instruction
Minimum:	78	128
First quartile:	117.50	159.75
Median:	147.50	191.50
Third quartile:	172.00	221.50
Maximum:	359	283
low extremes	0	0
low outliers	0	0
high outliers	2	0
high extremes	1	0

(c)

To determine

To find: the mean and standard deviation and explain whether it defines anything about shape or not.

Answer to Problem 2.6AYK

Group Instruction:

Mean = 163.28

Standard Deviation = 70.66

Individual instruction:

Mean =190.66

Standard Deviation = 41.90

Explanation of Solution

Formula used:

Mean = Average(“Array”)

Standard Deviation = stdev.s(“array”)

Calculation:

Statistics	Group Instruction	Individual instruction
Mean	163.28	190.06
Standard Deviation	70.66	41.90

No, mean and standard are the values that tells about the central value and the standard deviation tell about the spread of the data but not the shape of the data.

(d)

To determine

To represent the mean, one standard deviation above and one standard deviation below the mean on the dot plot of the given two datasets.

Explanation of Solution

The dot plots showing the mean and one standard deviation below and one standard deviation above than then mean is given below.