X=# of cigarettes smoked (0=none; light=1-10; heavy=11-20); y= Duration of stress test (min); N=25 0, 34 0, 32 0, 31 5, 27 8, 24 10, 23 15, 20 20, 21 20, 24 0, 25 0, 35 0, 26 4, 22 9, 22 6, 20 17, 15 20, 10 11, 13 0, 21 0, 20 0, 13 10, 17 8, 17 4, 14 10, 16
The given data shows the number of cigarettes smoked (X) and the duration of a stress test (Y) for a sample of 25 individuals. The variable X is categorized into three groups: none (0 cigarettes smoked), light (1-10 cigarettes smoked), and heavy (11-20 cigarettes smoked).
Let’s first organize the data in a more structured format:
X | Y
— | —
0 | 34
0 | 32
0 | 31
5 | 27
8 | 24
10 | 23
15 | 20
20 | 21
20 | 24
0 | 25
0 | 35
0 | 26
4 | 22
9 | 22
6 | 20
17 | 15
20 | 10
11 | 13
0 | 21
0 | 20
0 | 13
10 | 17
8 | 17
4 | 14
10 | 16
Now that we have the data organized, we can perform a descriptive analysis to understand the relationship between the number of cigarettes smoked and the duration of the stress test.
To start, we will calculate the mean, median, and mode for both X and Y.
Mean (X) = (0 + 0 + 0 + 5 + 8 + 10 + 15 + 20 + 20 + 0 + 0 + 0 + 4 + 9 + 6 + 17 + 20 + 11 + 0 + 0 + 0 + 10 + 8 + 4 + 10) / 25 = 5.28
Mean (Y) = (34 + 32 + 31 + 27 + 24 + 23 + 20 + 21 + 24 + 25 + 35 + 26 + 22 + 22 + 20 + 15 + 10 + 13 + 21 + 20 + 13 + 17 + 17 + 14 + 16) / 25 = 20.6
Median (X) = 10
Median (Y) = 21
Mode (X) = 0, as it appears most frequently
Mode (Y) = 20, as it appears most frequently
Next, we will calculate the standard deviation for both X and Y.
Standard deviation (X) = √((Σ (Xi – X̄)²) / (N – 1))
= √(((0 – 5.28)² + (0 – 5.28)² + (0 – 5.28)² + (5 – 5.28)² + (8 – 5.28)² + (10 – 5.28)² + (15 – 5.28)² + (20 – 5.28)² + (20 – 5.28)² + (0 – 5.28)² + (0 – 5.28)² + (0 – 5.28)² + (4 – 5.28)² + (9 – 5.28)² + (6 – 5.28)² + (17 – 5.28)² + (20 – 5.28)² + (11 – 5.28)² + (0 – 5.28)² + (0 – 5.28)² + (0 – 5.28)² + (10 – 5.28)² + (8 – 5.28)² + (4 – 5.28)² + (10 – 5.28)²)) / (25 – 1))
= √((683.52) / 24)
= √(28.48)
= 5.33
Standard deviation (Y) = √((Σ (Yi – Ȳ)²) / (N – 1))
= √(((34 – 20.6)² + (32 – 20.6)² + (31 – 20.6)² + (27 – 20.6)² + (24 – 20.6)² + (23 – 20.6)² + (20 – 20.6)² + (21 – 20.6)² + (24 – 20.6)² + (25 – 20.6)² + (35 – 20.6)² + (26 – 20.6)² + (22 – 20.6)² + (22 – 20.6)² + (20 – 20.6)² + (15 – 20.6)² + (10 – 20.6)² + (13 – 20.6)² + (21 – 20.6)² + (20 – 20.6)² + (13 – 20.6)² + (17 – 20.6)² + (17 – 20.6)² + (14 – 20.6)² + (16 – 20.6)²)) / (25 – 1))
= √((256.6) / 24)
= √(10.6916667)
= 3.27
The mean and median provide measures of central tendency, indicating the average and middle value of the data, respectively. In this case, the mean number of cigarettes smoked is approximately 5.28, while the mean duration of the stress test is approximately 20.6 minutes. The median values for both X and Y are 10 and 21, respectively. The mode represents the most frequently occurring value, which is 0 for X and 20 for Y.
The standard deviation provides a measure of dispersion, showing how spread out the data is around the mean. For X, the standard deviation is approximately 5.33, indicating that the values are relatively spread out. For Y, the standard deviation is approximately 3.27, suggesting less variability in the data.
From this initial analysis, we can see that there is variability in the duration of the stress test for different levels of cigarette smoking. The mean and median values for both X and Y indicate that there may be differences in the duration of the stress test based on the number of cigarettes smoked. However, to draw more definitive conclusions, further statistical analysis such as hypothesis testing or regression analysis would be required.