Question

blocking refers to the idea that the variability in a variable can be reduced by segmenting the data by some other variable. the data in the accompanying table represent the recumbent length (in centimeters) of a sample of 10 males and 10 females who are 40 months of age. complete parts (a) through (d).
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(a) determine the standard deviation of recumbent length for all 20 observations.
6.08 cm (round to two decimal places as needed.)
(b) determine the standard deviation of recumbent length for the males.
5.54 cm (round to two decimal places as needed.)
(c) determine the standard deviation of recumbent length for the females.
□ cm (round to two decimal places as needed.)

full data set
males females
103.0 94.4 | 102.5 100.8
93.7 97.6 | 100.0 96.3
98.3 100.6 | 102.8 105.0
86.2 103.0 | 98.1 106.5
90.7 100.9 | 95.4 114.5

Explanation:

Step1: Calculate female data mean

First, sum all female lengths and divide by 10.
Female data: 102.5, 100.8, 100.0, 96.3, 102.8, 105.0, 98.1, 106.5, 95.4, 114.5
Sum = $102.5 + 100.8 + 100.0 + 96.3 + 102.8 + 105.0 + 98.1 + 106.5 + 95.4 + 114.5 = 1021.9$
Mean $\bar{x} = \frac{1021.9}{10} = 102.19$

Step2: Compute squared deviations

Find $(x_i - \bar{x})^2$ for each value:
$(102.5-102.19)^2 = 0.0961$, $(100.8-102.19)^2 = 1.9321$, $(100.0-102.19)^2 = 4.7961$, $(96.3-102.19)^2 = 34.6921$, $(102.8-102.19)^2 = 0.3721$, $(105.0-102.19)^2 = 7.8961$, $(98.1-102.19)^2 = 16.7281$, $(106.5-102.19)^2 = 18.5761$, $(95.4-102.19)^2 = 46.1041$, $(114.5-102.19)^2 = 151.5361$

Step3: Sum squared deviations

Sum = $0.0961+1.9321+4.7961+34.6921+0.3721+7.8961+16.7281+18.5761+46.1041+151.5361 = 282.629$

Step4: Calculate sample variance

Variance $s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{282.629}{9} \approx 31.4032$

Step5: Find standard deviation

Standard deviation $s = \sqrt{31.4032} \approx 5.60$

Answer:

5.60 cm