High School

Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM. Find the values of:

a) [tex]\bar{d}[/tex]

b) [tex]s_d[/tex]

c) In general, what does [tex]\bar{d}[/tex] represent?

Temperatures at 8 AM: 98.3, 98.9, 97.2, 97.1, 97.4

Temperatures at 12 AM: 98.7, 99.3, 97.7, 96.7, 97.9

Answer :

a) d-bar (mean of differences): 0.4

b) sub d (standard deviation of differences): approximately 0.363

c) The mean d represents the average change or difference between the values in two sets of observations. In this specific case, it represents the average change in body temperature from 8 AM to 12 AM across the five subjects.

To find the values of d-bar and sub d, we first need to calculate the differences between the temperatures at 8 AM and 12 AM for each subject.

Temperatures at 8 AM: 98.3, 98.9, 97.2, 97.1, 97.4

Temperatures at 12 AM: 98.7, 99.3, 97.7, 96.7, 97.9

Calculating the differences:

d1 = (temperature at 12 AM) - (temperature at 8 AM)

d2 = (temperature at 12 AM) - (temperature at 8 AM)

d3 = (temperature at 12 AM) - (temperature at 8 AM)

d4 = (temperature at 12 AM) - (temperature at 8 AM)

d5 = (temperature at 12 AM) - (temperature at 8 AM)

d1 = 98.7 - 98.3 = 0.4

d2 = 99.3 - 98.9 = 0.4

d3 = 97.7 - 97.2 = 0.5

d4 = 96.7 - 97.1 = -0.4

d5 = 97.9 - 97.4 = 0.5

a) d-bar (mean of differences):

d-bar = (d1 + d2 + d3 + d4 + d5) / 5

= (0.4 + 0.4 + 0.5 - 0.4 + 0.5) / 5

= 0.4

b) sub d (standard deviation of differences):

To calculate the standard deviation of differences, we first find the squared differences and then take the square root of their average.

Squared differences:

[tex](d1 - d-bar)^2, (d2 - d-bar)^2, (d3 - d-bar)^2, (d4 - d-bar)^2, (d5 - d-bar)^2\\(d1 - d-bar)^2 = (0.4 - 0.4)^2 = 0\\(d2 - d-bar)^2 = (0.4 - 0.4)^2 = 0\\(d3 - d-bar)^2 = (0.5 - 0.4)^2 = 0.01\\(d4 - d-bar)^2 = (-0.4 - 0.4)^2 = 0.64\\(d5 - d-bar)^2 = (0.5 - 0.4)^2 = 0.01[/tex]

Calculating the average of squared differences:

[tex](sub d)^2 = [(d1 - d-bar)^2 + (d2 - d-bar)^2 + (d3 - d-bar)^2 + (d4 - d-bar)^2 + (d5 - d-bar)^2] / 5[/tex]

= (0 + 0 + 0.01 + 0.64 + 0.01) / 5

= 0.132

[tex]sub d = sqrt((sub d)^2)[/tex]

= sqrt(0.132)

≈ 0.363

c) In general, the mean d represents the average difference between the values in two sets of observations. In this case, it represents the average change in body temperature from 8 AM to 12 AM across the five subjects.

Learn more about deviation: https://brainly.in/question/4593664

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