Math Problem Statement

A medical researcher believes that a drug changes the body's temperature. Seven test subjects are randomly selected and the body temperature of each is measured. The subjects are then given the drug, and after 30

minutes, the body temperature of each is measured again. The results are listed in the table below. Is there enough evidence to conclude that the drug changes the body's temperature?

Let d=(body temperature after taking drug)−(body temperature before taking drug)

. Use a significance level of α=0.01

for the test. Assume that the body temperatures are normally distributed for the population of people both before and after taking the drug.

Subject

1

2

3

4

5

6

7

Temperature (before)

99.1

99.9

98.9

98.8

99.7

99.8

99.9

Temperature (after)

98.8

99.3

99.6

98.5

99.5

99.3

99.4

Step 2 of 5 :

Find the value of the standard deviation of the paired differences. Round your answer to two decimal places.

Solution

To find the standard deviation of the paired differences (dd), we first calculate the differences between the body temperatures before and after taking the drug. Then, we calculate the standard deviation of these differences.

Step 1: Calculate the differences (dd)

For each subject: d=Temperature (after)Temperature (before)d = \text{Temperature (after)} - \text{Temperature (before)}

d1=98.899.1=0.3d_1 = 98.8 - 99.1 = -0.3 d2=99.399.9=0.6d_2 = 99.3 - 99.9 = -0.6 d3=99.698.9=0.7d_3 = 99.6 - 98.9 = 0.7 d4=98.598.8=0.3d_4 = 98.5 - 98.8 = -0.3 d5=99.599.7=0.2d_5 = 99.5 - 99.7 = -0.2 d6=99.399.8=0.5d_6 = 99.3 - 99.8 = -0.5 d7=99.499.9=0.5d_7 = 99.4 - 99.9 = -0.5

The differences (dd) are: d=[0.3,0.6,0.7,0.3,0.2,0.5,0.5]d = [-0.3, -0.6, 0.7, -0.3, -0.2, -0.5, -0.5]

Step 2: Calculate the mean of dd

The mean of dd (dˉ\bar{d}) is: dˉ=dn=0.3+(0.6)+0.7+(0.3)+(0.2)+(0.5)+(0.5)7\bar{d} = \frac{\sum d}{n} = \frac{-0.3 + (-0.6) + 0.7 + (-0.3) + (-0.2) + (-0.5) + (-0.5)}{7} dˉ=1.77=0.2428570.24\bar{d} = \frac{-1.7}{7} = -0.242857 \approx -0.24

Step 3: Calculate the deviations from the mean

For each did_i, calculate (didˉ)2(d_i - \bar{d})^2: (d1dˉ)2=(0.3(0.24))2=(0.06)2=0.0036(d_1 - \bar{d})^2 = (-0.3 - (-0.24))^2 = (-0.06)^2 = 0.0036 (d2dˉ)2=(0.6(0.24))2=(0.36)2=0.1296(d_2 - \bar{d})^2 = (-0.6 - (-0.24))^2 = (-0.36)^2 = 0.1296 (d3dˉ)2=(0.7(0.24))2=(0.94)2=0.8836(d_3 - \bar{d})^2 = (0.7 - (-0.24))^2 = (0.94)^2 = 0.8836 (d4dˉ)2=(0.3(0.24))2=(0.06)2=0.0036(d_4 - \bar{d})^2 = (-0.3 - (-0.24))^2 = (-0.06)^2 = 0.0036 (d5dˉ)2=(0.2(0.24))2=(0.04)2=0.0016(d_5 - \bar{d})^2 = (-0.2 - (-0.24))^2 = (0.04)^2 = 0.0016 (d6dˉ)2=(0.5(0.24))2=(0.26)2=0.0676(d_6 - \bar{d})^2 = (-0.5 - (-0.24))^2 = (-0.26)^2 = 0.0676 (d7dˉ)2=(0.5(0.24))2=(0.26)2=0.0676(d_7 - \bar{d})^2 = (-0.5 - (-0.24))^2 = (-0.26)^2 = 0.0676

Step 4: Calculate the variance

The variance (s2s^2) of the differences is: s2=(didˉ)2n1s^2 = \frac{\sum (d_i - \bar{d})^2}{n-1} s2=0.0036+0.1296+0.8836+0.0036+0.0016+0.0676+0.067671s^2 = \frac{0.0036 + 0.1296 + 0.8836 + 0.0036 + 0.0016 + 0.0676 + 0.0676}{7-1} s2=1.157260.1929s^2 = \frac{1.1572}{6} \approx 0.1929

Step 5: Calculate the standard deviation

The standard deviation (ss) is the square root of the variance: s=0.19290.44s = \sqrt{0.1929} \approx 0.44

Final Answer:

The standard deviation of the paired differences is: 0.44\boxed{0.44}

Let me know if you'd like to proceed with additional steps in the hypothesis test or need further clarification!


Relative Questions:

  1. What is the next step in conducting the hypothesis test for paired differences?
  2. How do we determine the test statistic for this problem?
  3. What role does the significance level (α\alpha) play in this test?
  4. How do we interpret a result when pp-value < α\alpha?
  5. What assumptions must hold for a paired tt-test to be valid?

Tip: Always verify normality assumptions in paired tt-tests, especially for small sample sizes.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
Paired Sample Analysis
Standard Deviation

Formulas

d = (Temperature after taking drug) - (Temperature before taking drug)
Mean of differences (d̄) = Σd / n
Variance (s^2) = Σ(d_i - d̄)^2 / (n - 1)
Standard Deviation (s) = √s^2

Theorems

Paired Sample t-Test Assumptions

Suitable Grade Level

Grades 11-12, Undergraduate