High School

Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM.

- Temperature (∘F) at 8 AM: 98.5, 99.2, 97.4, 97.9, 97.4
- Temperature (∘F) at 12 AM: 99.2, 99.8, 97.7, 97.8, 97.5

Let the temperature at 8 AM be the first sample, and the temperature at 12 AM be the second sample. Find the values of [tex]\overline{d}[/tex] and [tex]s_d[/tex].

In general, what does [tex]\mu_d[/tex] represent?

A. The mean of the differences between each matched pair from the population of matched data.

Answer :

Mean difference (d¯) = 0.32 and standard deviation (sd) ≈ 0.3036.

To find the mean difference (d¯) and standard deviation (sd) of the paired data.

Calculate the differences between each pair of temperatures (12 AM - 8 AM).

Find the mean and standard deviation of these differences.

The differences for each pair:

1. 99.2 - 98.5 = 0.7

2. 99.8 - 99.2 = 0.6

3. 97.7 - 97.4 = 0.3

4. 97.8 - 97.9 = -0.1

5. 97.5 - 97.4 = 0.1

Mean difference (d¯) = (0.7 + 0.6 + 0.3 - 0.1 + 0.1) / 5

Mean difference (d¯) = 1.6 / 5

Mean difference (d¯) = 0.32

Calculate the squared differences from the mean difference:

(0.7 - 0.32)² = 0.1449

(0.6 - 0.32)² = 0.0784

(0.3 - 0.32)² = 0.0009

(-0.1 - 0.32)² = 0.1849

(0.1 - 0.32)² = 0.0516

Find the average of these squared differences:

Average squared difference = (0.1449 + 0.0784 + 0.0009 + 0.1849 + 0.0516) / 5 = 0.09214

Take the square root of the average squared difference to find the standard deviation (sd):

sd = √0.09214

sd ≈ 0.3036

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