Question

If we increase our food intake, we generally gain weight. Nutrition scientists can calculate the amount of weight gain that would be associated with a given increase in calories. In one study, 16 nonobese adults, aged 25 to 36 years, were fed 1000 calories per day in excess of the calories needed to maintain a stable body weight. The subjects maintained this diet for eight weeks, so they consumed a total of 56,000 extra calories.18 According to theory, 3500 extra calories will translate into a weight gain of 1 pound. Therefore, we expect each of these subjects to gain 56,000/3500 = 16 pounds (lb). Here are the weights before and after the 8-week period, expressed in kilograms (kg).
Subject 1 2 3 4 5 6 7 8Weight before 55.7 54.9 59.6 62.3 74.2 75.6 70.7 53.3Weight after 61.7 58.8 66.0 66.1 78.9 82.3 74.4 59.3Subject 9 10 11 12 13 14 15 16
Weight before 73.3 63.4 68.1 73.7 91.7 55.9 61.7 57.8Weight after 79.1 66.1 73.4 77.0 93.1 63.0 68.1 60.2(a) For each subject, subtract the weight before from the weight after to determine the weight change in kg.
(b) Find the mean and the standard deviation for the weight change. (Round your answers to four decimal places.)

Answer 1

Answer:

a) d: 6,3.9,6.4,3.8,4.7,6.7,3.7,6,5.8,2.7,5.3,3.3,1.4,7.1,6.4,2.4

b) $\bar d= \frac{\sum_{i=1}^n d_i}{n}=4.725$

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.733$

Step-by-step explanation:

x=test weight before , y = test weight after

x: 55.7, 54.9, 59.6, 62.3, 74.2 ,75.6, 70.7, 53.3,73.3, 63.4, 68.1, 73.7, 91.7, 55.9, 61.7, 57.8

y: 61.7, 58.8, 66.0, 66.1, 78.9, 82.3, 74.4, 59.3,79.1, 66.1, 73.4, 77.0, 93.1, 63.0, 68.1, 60.2

(a) For each subject, subtract the weight before from the weight after to determine the weight change in kg.

For this case we can define the difference $d_i=y_i-x_i$ and we obtain this:

Individual 1 : d1=61.7-55.7=6         Individual 2 : d2=58.8-54.9=3.9

Individual 3 : d3=66.0-59.6=6.4   Individual 4 : d4=66.1-62.3=3.8

Individual 5 : d5=78.9-74.2=4.7    Individual 6 : d6=82.3-75.6=6.7

Individual 7 : d7=74.4-70.7=3.7     Individual 8 : d8=59.3-53.3=6

Individual 9 : d9=79.1-73.3=5.8     Individual 10 : d10=66.1-63.4=2.7

Individual 11 : d11=73.4-68.1=5.3    Individual 12 : d12=77.0-73.7=3.3

Individual 13 : d13=93.1-91.7=1.4    Individual 14 : d14=63-55.9=7.1

Individual 15 : d15=68.1-61.7=6.4   Individual 16 : d16=60.2-57.8=2.4

(b) Find the mean and the standard deviation for the weight change. (Round your answers to four decimal places.)

We can calculate the mean difference  like this:

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=4.725$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =1.733$