Question

which is best at reducing stress? 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 effects of pets and friends in stressful situations, researchers recruited 45 people who said they were dog lovers. fifteen people 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 persons average heart rate during the task was one measure of the effect of stress. the following table shows the data: 101

	62.6	70.9	73.3	75.5	77.8	80.4	84.5	84.7
	76.9	80.3	81.6	83.4	87.0	88.0	89.8	91.4
friend	92.5	97.0	98.2	99.7	100.9	101.1	102.2
	58.7	64.2	65.4	68.9	69.2	69.2	69.5	70.1
pet	70.2	72.3	76.0	79.7	85.0	86.4	97.5

1. show that the value 97.5 is an outlier in the pet distribution. (note that there are no other outliers in these three distributions.)
2. make parallel boxplots to compare the heart rates of the people 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? explain your answer.

Explanation:

Step1: Calculate quartiles for pet - group

First, order the pet - group 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. There are \(n = 15\) data points. The median (\(Q_2\)) is the 8th value, so \(Q_2=70.1\). The lower half of the data is 58.7, 64.2, 65.4, 68.9, 69.2, 69.2, 69.5. The median of the lower half (\(Q_1\)) is the 4th value, so \(Q_1 = 68.9\). The upper half of the data is 70.2, 72.3, 76.0, 79.7, 85.0, 86.4, 97.5. The median of the upper half (\(Q_3\)) is the 4th value of the upper - half, so \(Q_3=79.7\).

Step2: Calculate the inter - quartile range (IQR)

\(IQR=Q_3 - Q_1=79.7−68.9 = 10.8\).

Step3: Determine the lower and upper fences

The lower fence is \(Q_1-1.5\times IQR=68.9−1.5\times10.8=68.9 - 16.2 = 52.7\). The upper fence is \(Q_3 + 1.5\times IQR=79.7+1.5\times10.8=79.7 + 16.2=95.9\). Since \(97.5>95.9\), 97.5 is an outlier in the pet distribution.

Step4: For parallel boxplots

To make parallel boxplots, we need to find the five - number summary (minimum, \(Q_1\), \(Q_2\), \(Q_3\), maximum) for each group:

• Alone group: Minimum = 62.6, \(Q_1\) (4th value of ordered data) = 75.5, \(Q_2\) (8th value) = 84.7, \(Q_3\) (12th value) = 87.8, Maximum = 99.0.
• Friend group: Minimum = 76.9, \(Q_1\) (4th value) = 83.4, \(Q_2\) (8th value) = 91.4, \(Q_3\) (12th value) = 99.7, Maximum = 102.2.
• Pet group: Minimum = 58.7, \(Q_1\) = 68.9, \(Q_2\) = 70.1, \(Q_3\) = 79.7, Maximum = 97.5 (with outlier). Then we can draw the boxplots with the same scale on the x - axis for each group.

Step5: Analyze the effect on heart rate

The median heart rate for the alone group is 84.7, for the friend group is 91.4, and for the pet group is 70.1 (excluding the outlier 97.5). The mean and median heart rates for the pet group are generally lower than those of the alone and friend groups. This suggests that the presence of a pet may reduce heart rate during a stressful task, while the presence of a friend does not seem to have a consistent reducing effect as the heart rate values in the friend group are relatively high compared to the pet group.

Answer:

1. The value 97.5 is an outlier in the pet distribution as it is greater than the upper fence of 95.9 calculated using \(Q_1 = 68.9\), \(Q_3=79.7\) and \(IQR = 10.8\).
2. Parallel boxplots can be drawn using the five - number summaries for each group (minimum, \(Q_1\), \(Q_2\), \(Q_3\), maximum) as calculated above.
3. The presence of a pet may reduce heart rate during a stressful task as the median and mean (excluding the outlier) heart rates for the pet group are lower than those of the alone and friend groups, while the presence of a friend does not seem to have a consistent reducing effect.