Math Problem Statement

To test the effect of a physical fitness course on one's physical ability, the number of sit-ups that a person could do in one minute, both before and after the course, was recorded. Ten individuals are randomly selected to participate in the course. The results are displayed in the following table. Using this data, find the 95% confidence interval for the true difference in the number of sit-ups each person can do before and after the course. Assume that the numbers of sit-ups are normally distributed for the population both before and after completing the course.

Sit-ups before 52 22 35 31 53 35 47 29 22 44 Sit-ups after 57 33 42 46 56 46 53 46 38 49

Step 2 of 4 : Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.

Solution

To find the sample standard deviation of the paired differences between sit-ups before and after the course, follow these steps:

  1. Calculate the differences for each individual: Difference=Sit-ups After−Sit-ups Before\text{Difference} = \text{Sit-ups After} - \text{Sit-ups Before}

    Using the provided dataTo find the sample standard deviation of the paired differences between sit-ups before and after the course, follow these steps:

  2. Calculate the differences for each individual: Difference=Sit-ups After−Sit-ups Before\text{Difference} = \text{Sit-ups After} - \text{Sit-ups Before}

    Using the provided

    • Differences:
      • 57−52=557 - 52 = 5
      • 33−22=1133 - 22 = 11
      • 42−35=742 - 35 = 7
      • 46−31=1546 - 31 = 15
      • 56−53=356 - 53 = 3
      • 46−35=1146 - 35 = 11
      • 53−47=653 - 47 = 6
      • 46−29=1746 - 29 = 17
      • 38−22=1638 - 22 = 16
      • 49−44=549 - 44 = 5

    Thus, the differences are: 5,11,7,15,3,11,6,17,16,55, 11, 7, 15, 3, 11, 6, 17, 16, 5

  3. Calculate the mean of the differences (dˉ\bar{d}): dˉ=5+11+7+15+3+11+6+17+16+510=5+11+7+15+3+11+6+17+16+510=11+11+7+15+3+11+6+17+16+510=5+11+11+15+3+11+6+17+16+510=10010=10\bar{d} = \frac{5 + 11 + 7 + 15 + 3 + 11 + 6 + 17 + 16 + 5}{10} = \frac{ 5 + 11 + 7 + 15 + 3 + 11 + 6 + 17 + 16 + 5}{10} = \frac{ 11 + 11 + 7 + 15 + 3 + 11 + 6 + 17 + 16 + 5}{10} = \frac{ 5 + 11 + 11 + 15 + 3 + 11 + 6 + 17 + 16 + 5}{10} = \frac{ 100}{10} = 10

  4. Calculate the squared differences from the mean: (di−dˉ)2(d_i - \bar{d})^2

    • For 5:(5−10)2=255: (5 - 10)^2 = 25
    • For 11:(11−10)2=111: (11 - 10)^2 = 1
    • For 7:(7−10)2=97: (7 - 10)^2 = 9
    • For 15:(15−10)2=2515: (15 - 10)^2 = 25
    • For 3:(3−10)2=493: (3 - 10)^2 = 49
    • For 11:(11−10)2=111: (11 - 10)^2 = 1
    • For 6:(6−10)2=166: (6 - 10)^2 = 16
    • For 17:(17−10)2=4917: (17 - 10)^2 = 49
    • For 16:(16−10)2=3616: (16 - 10)^2 = 36
    • For 5:(5−10)2=255: (5 - 10)^2 = 25

  5. Sum the squared differences: 25+1+9+25+49+1+16+49+36+25=23625 + 1 + 9 + 25 + 49 + 1 + 16 + 49 + 36 + 25 = 236

  6. Calculate the sample variance: s2=Sum of squared differencesn−1=23610−1=2369≈26.222222s^2 = \frac{\text{Sum of squared differences}}{n - 1} = \frac{236}{10 - 1} = \frac{236}{9} \approx 26.222222

  7. Calculate the sample standard deviation (ss): s=26.222222≈5.121s = \sqrt{26.222222} \approx 5.121

Rounding to six decimal places, the sample standard deviation of the paired differences is: s≈5.121000s \approx 5.121000

Summary:

The sample standard deviation of the paired differences is approximately 5.121000.

Would you like more details or have any questions about this calculation? Here are some follow-up questions:

  1. How would you calculate the mean of the paired differences?
  2. What is the next step to find the 95% confidence interval?
  3. Can you explain the significance of the paired t-test?
  4. How would the confidence interval change if the sample size increased?
  5. What assumptions are necessary for this analysis?

Tip: Always ensure your data meets the assumptions of normality when performing parametric tests like the paired t-test.

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Math Problem Analysis

Mathematical Concepts

Statistics
Paired t-test
Confidence Intervals

Formulas

Difference = Sit-ups After - Sit-ups Before
Mean of differences (d̄) = ΣD/n
Sample variance (s²) = Σ(D - d̄)² / (n - 1)
Sample standard deviation (s) = √s²

Theorems

Central Limit Theorem
Paired Samples t-test

Suitable Grade Level

Grades 11-12

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