Question

For each variable of interest, do the following: 1. Find the mean, five-number summary, range, variance, and standard deviation. Display these numbers in a format that is easy to understand. 2. For each variable of interest, use its five-number summary to construct a boxplot. Each boxplot must be constructed horizontally, and must be accompanied by a brief descriptive paragraph that assesses whether the data appear to be symmetrical, left-skewed, or right-skewed.

Construct a 95% confidence interval for the mean μ of each variable of interest (two confidence intervals total).

Pct Time Asleep	Longevity
22	35
9	37
49	49
1	46
23	63
83	39
23	46
15	56
9	63
81	65
12	56
15	65
37	70
24	63
26	65
17	70
14	77
14	81
6	86
25	70
18	70
26	77
24	77
29	81
27	77
18	40
6	37
19	44
7	47
16	47
13	47
35	68
2	47
35	54
6	61
15	71
14	75
18	89
50	58
25	59
10	62
33	79
43	96
35	58
17	62
27	70
22	72
16	75
20	96
37	75
23	46
4	42
20	65
42	46
9	58
32	42
66	48
28	58
10	50
4	80
12	63
17	65
12	70
23	70
40	72
18	97
10	46
38	56
7	70
23	70
36	72
9	76
21	90
62	76
36	92
23	21
62	40
28	44
18	54
10	36
28	40
22	56
29	60
15	48
73	53
10	60
5	60
13	65
27	68
20	60
21	81
12	81
49	48
17	48
22	56
71	68
17	75
10	81
24	48
18	68
34	16
6	19
4	19
22	32
28	33
31	33
16	30
27	42
8	42
32	33
20	26
35	30
12	40
14	54
17	34
29	34
31	47
6	47
30	42
27	47
40	54
19	54
8	56
8	60
15	44

Answer 1

Here, we take the variable Pct Time Asleep as X and Longevity as Y.

1.

The following table shows the mean, five-number summary (i.e. minimum, maximum, first quartile Q1, median Q2, third quartile Q3), range, variance and standard deviation.

Variables	Mean	StDev	Variance	Minimum	Q1	Median	Q3	Maximum	Range
Pct Time Asleep (X)	1.42	15.88	252.14	1	12.5	20	29	83	82
Longevity (Y)	1.57	17.56	308.49	16	46	58	70	97	81

2.

Data Pct Time Asleep (X) Longevity (Y) 0 20 40 60 80 100

For the Pct Time Asleep (X) Variable, the minimum value (1), Maximum value (83), First Quartile Q1 (12.5), Median Q2 (20) and Third Quartile Q3 (29). So, with the help of this data we draw the box plot for Pct time Asleep (X) Variable.

For the Longevity (Y) Variable, the minimum value (16), Maximum value (97), First Quartile Q1 (46), Median Q2 (58) and Third Quartile Q3 (70). So, with the help of this data we draw the box plot for Longevity (Y) Variable.

Here, for the Variable Pct Time Asleep (X), the distance (9 unit) between the upper hinge (29) & Median (20) is greater than the distance (7.5 unit) between the the lower hinge (12.5) & Median (20). Hence, the data for the variable Pct Time Asleep (X) is positively skewed i.e. it is a right tail or right-skewed.

Here, for the Variable Longevity (Y), the distance (12 unit) between the upper hinge (70) & Median (58) is equal to the distance (12 unit) between the the lower hinge (46) & Median (58). Hence, the data for the variable Longevity (Y) is symmetrical.

Answer: The data for the variable Pct Time Asleep (X) is right-skewed and the data for the variable Longevity (Y) is symmetrical.

Here, for the Variable Pct Time Asleep (X), 7 values are outliers i.e. 74 th value (62), 77 th value (62), 57th value (66), 96th value (71), 85th value (73), 10th value (81) and 6th value (83). These are outliers.

But, for the Variable Longevity (Y), there is no outliers.

95% confidence interval for the mean $\mu$ is calculated by this formula:

$\bar{x} \pm (Z_{\alpha } *\frac{sd}{\sqrt{n}})$

Now, for the Variable Pct Time Asleep (X), the CI is

$1.42 \pm (1.96 *\frac{15.88}{\sqrt{125}})$

= (1.42 - 2.78, 1.42 + 2.78)

= (-1.36, 4.2)

Now, for the Variable Longevity (Y), the CI is

$1.57 \pm (1.96 *\frac{17.56}{\sqrt{125}})$

= (1.57 - 3.08, 1.57 + 3.08)

= (-1.51, 4.65)

Answer: 95% confidence interval for the mean $\mu$ for the Variable Pct Time Asleep (X), is (-1.36, 4.2) & for the Variable Longevity (Y) is (-1.51, 4.65).