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Solved: Blocking and Variability Recall that blocking
Chapter 9, Problem 41AYU(choose chapter or problem)
Blocking refers to the idea that we can reduce the variability in a variable by segmenting the data by some other variable. The given data represent the recumbent length (in centimeters) of a sample of 10 males and 10 females who are 40 months of age.
MALES |
FEMALES | ||
104.0 |
94.4 |
102.5 |
100.8 |
|---|---|---|---|
93.7 |
97.6 |
100.4 |
96.3 |
98.3 |
100.6 |
102.7 |
105.0 |
86.2 |
103.0 |
98.1 |
106.5 |
90.7 |
100.9 |
95.4 |
114.5 |
(a) Determine the standard deviation of recumbent length for all 20 observations.
(b) Determine the standard deviation of recumbent length for the males.
(c) Determine the standard deviation of recumbent length for the females.
(d) What effect does blocking by gender have on the standard deviation of recumbent length for each gender?
Questions & Answers
QUESTION:
Blocking refers to the idea that we can reduce the variability in a variable by segmenting the data by some other variable. The given data represent the recumbent length (in centimeters) of a sample of 10 males and 10 females who are 40 months of age.
MALES |
FEMALES | ||
104.0 |
94.4 |
102.5 |
100.8 |
|---|---|---|---|
93.7 |
97.6 |
100.4 |
96.3 |
98.3 |
100.6 |
102.7 |
105.0 |
86.2 |
103.0 |
98.1 |
106.5 |
90.7 |
100.9 |
95.4 |
114.5 |
(a) Determine the standard deviation of recumbent length for all 20 observations.
(b) Determine the standard deviation of recumbent length for the males.
(c) Determine the standard deviation of recumbent length for the females.
(d) What effect does blocking by gender have on the standard deviation of recumbent length for each gender?
ANSWER:
Step 1 of 4
a)
Consider the data.
|
MALES |
FEMALES |
||
|
104.0 |
94.4 |
102.5 |
100.8 |
|
93.7 |
97.6 |
100.4 |
96.3 |
|
98.3 |
100.6 |
102.7 |
105.0 |
|
86.2 |
103.0 |
98.1 |
106.5 |
|
90.7 |
100.9 |
95.4 |
114.5 |
Here the standard deviation of recumbent length for all 20 observations.
So, n = 20.
First we need to find the sample variance.
mean = \(=\frac{\sum_{i=1}^{n} x_{i}}{n}\)
mean = \(\frac{1991.6}{20}\)
mean = \(99.58\)
Then the standard deviation is
\(s^{2}=\frac{\Sigma\left(x_{i}-\bar{x}\right)^{2}}{n-1}\)
\(s^{2}=\frac{19.5364+34.5744+\ldots+222.6064}{20-1}\)
\(s^{2}=\frac{709.572}{19}\)
\(s^{2}=\mathbf{3 7 . 3 4 5 8}\)
Then the sample variance is
\(\mathrm{s}=\sqrt{\frac{\Sigma\left(x_{i}-\bar{x}\right)^{2}}{n-1}}\)
\(\mathrm{s}=\sqrt{37.3458}\)
\(\mathrm{s}=6.11\)
Hence the sample variance is \(\boxed{\mathrm{s}=6.11\ cm}\).