Math Problem Statement

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Part 1

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

d overbard

and

s Subscript dsd.

In​ general, what does

mu Subscript dμd

​represent?

Temperature left parenthesis degrees Upper F right parenthesis at 8 AMTemperature (°F) at 8 AM

98.498.4

98.998.9

97.597.5

97.397.3

97.797.7   Temperature left parenthesis degrees Upper F right parenthesis at 12 AMTemperature (°F) at 12 AM

99.099.0

99.299.2

97.797.7

96.996.9

98.098.0

Question content area bottom

Part 1

Let the temperature at 8 AM be the first​ sample, and the temperature at 12 AM be the second sample. Find the values of

d overbard

and

s Subscript dsd.

d overbardequals=enter your response here

​(Type an integer or a decimal. Do not​ round.)

Solution

To calculate d\overline{d} (the mean of the differences) and sds_d (the standard deviation of the differences), we proceed as follows:

  1. Calculate the differences (dd) for each pair of temperatures: d=Temperature at 12 AMTemperature at 8 AMd = \text{Temperature at 12 AM} - \text{Temperature at 8 AM}.

  2. Calculate the mean difference (d\overline{d}): d=dn\overline{d} = \frac{\sum d}{n} where nn is the number of differences.

  3. Calculate the standard deviation (sds_d): sd=(dd)2n1s_d = \sqrt{\frac{\sum (d - \overline{d})^2}{n - 1}}

Step-by-step Calculation:

The data is:

Temperature (°F) at 8 AMTemperature (°F) at 12 AMd=12 AM8 AMd = \text{12 AM} - \text{8 AM}
98.499.099.098.4=0.699.0 - 98.4 = 0.6
98.999.299.298.9=0.399.2 - 98.9 = 0.3
97.597.797.797.5=0.297.7 - 97.5 = 0.2
97.396.996.997.3=0.496.9 - 97.3 = -0.4
97.798.098.097.7=0.398.0 - 97.7 = 0.3

The differences (dd) are: d=[0.6,0.3,0.2,0.4,0.3]d = [0.6, 0.3, 0.2, -0.4, 0.3]

1. Calculate d\overline{d}:

d=dn=0.6+0.3+0.20.4+0.35=1.05=0.2\overline{d} = \frac{\sum d}{n} = \frac{0.6 + 0.3 + 0.2 - 0.4 + 0.3}{5} = \frac{1.0}{5} = 0.2

2. Calculate sds_d:

First, compute (dd)2(d - \overline{d})^2 for each difference:

[ d - \overline{d} = [0.6 - 0.2, 0.3 - 0.2, 0.2 - 0.2, -0.4 - 0.2, 0.3 - 0.2] = [0.4, 0.1, 0.0, -0.6, 0.1] ]

(dd)2=[0.16,0.01,0.00,0.36,0.01](d - \overline{d})^2 = [0.16, 0.01, 0.00, 0.36, 0.01]

Sum of squares: (dd)2=0.16+0.01+0.00+0.36+0.01=0.54\sum (d - \overline{d})^2 = 0.16 + 0.01 + 0.00 + 0.36 + 0.01 = 0.54

Now calculate sds_d: sd=(dd)2n1=0.5451=0.544=0.1350.367s_d = \sqrt{\frac{\sum (d - \overline{d})^2}{n - 1}} = \sqrt{\frac{0.54}{5 - 1}} = \sqrt{\frac{0.54}{4}} = \sqrt{0.135} \approx 0.367

Final Results:

  • d=0.2\overline{d} = 0.2
  • sd0.367s_d \approx 0.367

Interpretation of μd\mu_d:

The parameter μd\mu_d represents the population mean of the differences between the temperatures at 12 AM and 8 AM. This is the average difference we would expect if we measured all such pairs in the population.


Let me know if you would like further clarification or if you have additional questions!

Expanding Questions:

  1. How is the standard deviation sds_d used in hypothesis testing for paired samples?
  2. Why is the denominator n1n - 1 used instead of nn in the standard deviation formula?
  3. Can the sign of dd indicate a systematic temperature change between the times?
  4. How would adding more measurements affect d\overline{d} and sds_d?
  5. How could we test if the difference between times is statistically significant?

Tip:

When calculating differences in paired data, always ensure to define the subtraction order clearly to maintain consistency.

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Math Problem Analysis

Mathematical Concepts

Statistics
Paired Sample Analysis
Descriptive Statistics

Formulas

Mean of differences: \( \overline{d} = \frac{\sum d}{n} \)
Standard deviation of differences: \( s_d = \sqrt{\frac{\sum (d - \overline{d})^2}{n - 1}} \)

Theorems

Descriptive statistics formulas for paired data

Suitable Grade Level

Grades 10-12