Serum ferritin is used in diagnosing iron deficiency. In a study conducted recently researchers discovered that in a sample of 28 elderly men the sample standard deviation of serum ferritin was 52.6 mg/L. For 26 younger men the sample standard deviation was 84.2 mg/L. At the .01 level of significance, do these data support the conclusion that the ferritin distribution in elderly men has a smaller variance than in younger men? Solution Given elderly: n1=28, s1=52.6 younger: n2=26, s2=84.2 The test hypothesis is Ho:a1^2 > a2^2 Ha:a1^2 < a2^2 The test statistic is F=s1^2/s2^2 = 52.6^2/ 84.2 ^2 = 0.39 Given aplha = 0.01, the critical value is F(0.99, d1=n2-1=25, d2=n1-1=27)= 2.54 (check F table) Since F=0.39 <2.54, we reject Ho. Answer: Yes, because the test value 0.390 is less than the critical value 2.54.

1. Serum zinc levels for 462 males between the ages of 15 and 17 are given below.
Use Chebychev's inequality to describe the distribution of values. Does the empirical rule do a
better job of summarizing these serum levels than Chebychev's inequality?
50
51
53
55
56
58
60
60
60
61
61
61
61
62
62
62
62
63
63
63
64
64
64
64
64
65
65
65

2. 66
66
66
67
67
67
67
67
68
68
68
68
69
70
70
70
70
70
70
70
70
70
71
71
71
71
71
71
71
71
71
71
71
72
72
72

3. 72
72
72
72
72
72
72
72
72
72
73
73
73
73
73
73
73
73
73
74
74
74
74
74
74
74
74
74
74
74
74
75
75
75
75
75

4. 75
75
75
75
75
75
75
75
75
75
76
76
76
76
76
76
76
76
76
76
76
76
76
77
77
77
77
77
77
77
77
77
77
78
78
78

5. 78
78
78
78
78
78
78
78
78
79
79
79
79
79
79
80
80
80
80
80
80
80
80
80
80
80
81
81
81
81
81
81
81
81
81
81

6. 82
82
82
82
82
82
82
82
82
82
82
82
82
83
83
83
83
83
83
83
83
83
83
83
84
84
84
84
84
84
84
84
84
84
84
84

7. 84
84
85
85
85
85
85
85
85
85
85
85
85
86
86
86
86
86
86
86
86
86
86
86
86
86
87
87
87
87
87
87
87
87
87
87

8. 88
88
88
88
88
88
88
88
88
88
88
88
88
88
89
89
89
89
89
89
89
89
89
90
90
90
90
90
90
90
90
91
91
91
91
91

9. 91
91
91
91
91
92
92
92
92
92
92
92
92
93
93
93
93
93
93
93
93
93
94
94
94
94
94
94
94
94
94
94
94
95
95
95

10. 95
95
95
95
95
95
95
95
96
96
96
96
96
96
96
96
96
96
97
97
97
97
97
97
97
97
97
98
98
98
98
98
98
98
98
98

11. 99
99
99
99
99
99
100
100
100
100
100
101
101
101
101
101
101
101
102
102
102
102
102
102
102
102
103
103
103
103
103
103
104
104
104
104

12. 104
104
104
105
105
105
105
105
105
105
105
106
106
106
106
106
106
107
107
107
107
107
107
107
108
108
108
108
108
108
108
109
109
110
110
111

13. 111
111
111
112
112
112
113
113
113
114
114
114
114
115
115
115
116
116
116
116
117
117
117
118
118
119
119
121
123
124
125
128
131
135
142
147

14. 151
153
I know what Chebychev's inequality is and how to explain what it is, I am just unsure how to
relate and apply it to the data.
Solution
a) Mean = xi/n = 87.94
Median = value in the middle = 86
Standard deviation = (xi-xbar)2/(n-1) = 16
Range = Max - Min = 103
Interquartile range of the data:
1st Quartile = 76
Median Value = 86
3rd Quartile = 98
Maximum value = 153
1st: 50 to 76
2nd: 77 to 86
3rd: 87 to 98
4th: 99 to 153