Question

if you are a dog lover, having your dog with you may reduce your stress level. does having a friend with you reduce stress? to examine the effect of pets and friends in stressful situations, researchers recruited 45 women who said they were dog lovers. fifteen women were assigned at random to each of three groups: to do a stressful task (1) alone, (2) with a good friend present, or (3) with their dogs present. the stressful task was to count backward by 13s or 17s. the woman’s average heart rate during the task was one measure of the effect of stress. the table shows the data.

alone: 62.6, 70.9, 73.3, 75.5, 77.8, 80.4, 84.5, 84.7, 84.9, 87.2, 87.4, 87.8, 90.0, 91.8, 99.0
friend: 76.9, 80.3, 81.6, 83.4, 87.0, 88.0, 89.8, 91.4, 92.5, 97.0, 98.2, 90.7, 100.9, 101.1, 102.2
pet: 58.7, 64.2, 65.4, 68.9, 69.2, 69.2, 69.5, 70.1, 70.2, 72.3, 76.0, 79.7, 85.0, 86.4, 97.5

1. identify any outliers in the three groups. show your work.
2. make parallel boxplots to compare the heart rates of the women in the three groups.
3. based on the data, does it appear that the presence of a pet or friend reduces heart rate during a stressful task? justify your answer.

Answer:

Question 1: Identify Outliers

Group: Alone

• Data: \( 62.6, 70.9, 73.3, 75.5, 77.8, 80.4, 84.5, 84.7, 84.9, 87.2, 87.4, 87.8, 90.0, 91.8, 99.0 \)
• Step 1: Find Quartiles
• Order data: \( 62.6, 70.9, 73.3, 75.5, 77.8, 80.4, 84.5, 84.7, 84.9, 87.2, 87.4, 87.8, 90.0, 91.8, 99.0 \)
• \( n = 15 \), so \( Q_1 \) (25th percentile) is at position \( \frac{15+1}{4} = 4 \): \( Q_1 = 75.5 \)
• \( Q_3 \) (75th percentile) is at position \( \frac{3(15+1)}{4} = 12 \): \( Q_3 = 87.8 \)
• Step 2: Calculate IQR

\( \text{IQR} = Q_3 - Q_1 = 87.8 - 75.5 = 12.3 \)

• Step 3: Outlier Bounds

Lower bound: \( Q_1 - 1.5 \times \text{IQR} = 75.5 - 18.45 = 57.05 \)
Upper bound: \( Q_3 + 1.5 \times \text{IQR} = 87.8 + 18.45 = 106.25 \)

• Step 4: Check Outliers

All data points are between \( 57.05 \) and \( 106.25 \). No outliers.

Group: Friend

• Data: \( 76.9, 80.3, 81.6, 83.4, 87.0, 88.0, 89.8, 91.4, 92.5, 97.0, 98.2, 90.7, 100.9, 101.1, 102.2 \) (Note: Data may have a typo; assume sorted: \( 76.9, 80.3, 81.6, 83.4, 87.0, 88.0, 89.8, 90.7, 91.4, 92.5, 97.0, 98.2, 100.9, 101.1, 102.2 \))
• Step 1: Find Quartiles
• \( n = 15 \), \( Q_1 \) at position \( 4 \): \( Q_1 = 83.4 \)
• \( Q_3 \) at position \( 12 \): \( Q_3 = 98.2 \)
• Step 2: Calculate IQR

\( \text{IQR} = 98.2 - 83.4 = 14.8 \)

• Step 3: Outlier Bounds

Lower bound: \( 83.4 - 22.2 = 61.2 \)
Upper bound: \( 98.2 + 22.2 = 120.4 \)

• Step 4: Check Outliers

All data points are within bounds. No outliers.

Group: Pet

• Data: \( 58.7, 64.2, 65.4, 68.9, 69.2, 69.2, 69.5, 70.1, 70.2, 72.3, 76.0, 79.7, 85.0, 86.4, 97.5 \)
• Step 1: Find Quartiles
• \( n = 15 \), \( Q_1 \) at position \( 4 \): \( Q_1 = 68.9 \)
• \( Q_3 \) at position \( 12 \): \( Q_3 = 79.7 \)
• Step 2: Calculate IQR

\( \text{IQR} = 79.7 - 68.9 = 10.8 \)

• Step 3: Outlier Bounds

Lower bound: \( 68.9 - 16.2 = 52.7 \)
Upper bound: \( 79.7 + 16.2 = 95.9 \)

• Step 4: Check Outliers

\( 97.5 > 95.9 \), so \( 97.5 \) is an outlier.

Question 2: Parallel Boxplots

To create parallel boxplots:

1. Axes: Horizontal axis (x) for groups (Alone, Friend, Pet); vertical axis (y) for heart rate.
2. Boxes: For each group, draw a box from \( Q_1 \) to \( Q_3 \), with a line at the median.
3. Whiskers: Extend from the box to the minimum/maximum (excluding outliers, which are plotted as dots).

• Alone: Median ≈ 84.7, \( Q_1 = 75.5 \), \( Q_3 = 87.8 \), min = 62.6, max = 99.0 (no outliers).
• Friend: Median ≈ 90.7, \( Q_1 = 83.4 \), \( Q_3 = 98.2 \), min = 76.9, max = 102.2 (no outliers).
• Pet: Median ≈ 70.1, \( Q_1 = 68.9 \), \( Q_3 = 79.7 \), min = 58.7, max = 86.4 (outlier at 97.5).

Question 3: Effect of Pet/Friend on Heart Rate

• Comparison:
• The "Alone" group has a higher median heart rate (≈84.7) than "Friend" (≈90.7? Wait, no—wait, recheck medians:
• Alone median (8th value): \( 84.7 \)
• Friend median (8th value): \( 90.7 \) (Wait, no—sorted Friend data: \( 76.9, 80.3, 81.6, 83.4, 87.0, 88.0, 89.8, 90.7, 91.4, 92.5, 97.0, 98.2, 100.9, 101.1, 102.2 \) → median is 90.7 (8th value).
• Pet median (8th value): \( 70.1 \) (sorted: \( 58.7, 64.2, 65.4, 68.9, 69.2, 69.2, 69.5, 70.1, ... \))

Wait, correction: The "Alone" group’s data includes higher values (e.g., 99.0), but the "Pet" group has a lower median (≈70) and "Friend" is mid (≈90). Wait, no—heart rate is a stress measure: lower heart rate = less stress.

• Pet Group: Lower median (≈70) and lower spread (IQR = 10.8) than "Alone" (IQR = 12.3) and "Friend" (IQR = 14.8).
• Friend Group: Median (≈90) is lower than "Alone" (≈84.7? No, 90 > 84.7—wait, no, "Alone" data: \( 62.6, 70.9, 73.3, 75.5, 77.8, 80.4, 84.5, 84.7, 84.9, 87.2, 87.4, 87.8, 90.0, 91.8, 99.0 \) → median is 84.7 (8th value). "Friend" median is 90.7 (higher than 84.7). "Pet" median is 70.1 (lower than 84.7).

Thus, the Pet group has a lower median heart rate, and the "Friend" group’s median is higher than "Alone"? Wait, maybe a miscalculation. Wait, "Alone" data: 15 values, 8th is 84.7. "Friend" data: 15 values, 8th is 90.7 (higher). "Pet" data: 8th is 70.1 (lower).

• Conclusion: The presence of a pet reduces heart rate (lower median, less spread) compared to being alone. The friend group’s median is higher than alone, suggesting friends may not reduce stress (or the task with friends had higher stress). But the pet group clearly has lower heart rates, indicating pets reduce stress.

Final Answers

1. Outliers:

• Alone: None
• Friend: None
• Pet: \( 97.5 \)

1. Parallel Boxplots: Drawn as described (boxes for each group with medians, whiskers, and outlier for Pet).

1. Effect: Yes, the presence of a pet reduces heart rate (lower median, less spread). The friend group’s median is higher than alone, but the pet group’s median is much lower, indicating pets reduce stress.