Dear Peter, thank you for this video and great explanation. May I kindly ask you to provide more details where the consistency factor 4.45 in the formulas for bounds for the detection of outliers comes from. Maybe you can provide a link to the literature or give us more details. Thank you in advance.
Found a possible explanation here, sharing if it helps. www.real-statistics.com/descriptive-statistics/mad-and-outliers/ "Another approach is to use the interval Median ± c ⋅ MAD where MAD is the median absolute deviation. We first note that for normally distributed data, μ = Median and σ ≈ 1.4826 ⋅ MAD. Here, 1.4826 = 1/NORM.S.INV(.75). Thus, for a normal distribution, the μ ± k ⋅ σ interval becomes Median ± 1.4826k ⋅ MAD. Thus, k = 2.5 for the normal distribution becomes c = 2.5 ⋅ 1.4826 = 3.7065 for this non-parametric version, and k = 3.0 for the normal distribution becomes c = 3.0 ⋅ 1.4826 = 4.4478."
Thanks for explaining this in an easy and practical way😊
Dear Peter, thank you for this video and great explanation. May I kindly ask you to provide more details where the consistency factor 4.45 in the formulas for bounds for the detection of outliers comes from. Maybe you can provide a link to the literature or give us more details. Thank you in advance.
Found a possible explanation here, sharing if it helps.
www.real-statistics.com/descriptive-statistics/mad-and-outliers/
"Another approach is to use the interval Median ± c ⋅ MAD where MAD is the median absolute deviation. We first note that for normally distributed data, μ = Median and σ ≈ 1.4826 ⋅ MAD. Here, 1.4826 = 1/NORM.S.INV(.75). Thus, for a normal distribution, the μ ± k ⋅ σ interval becomes Median ± 1.4826k ⋅ MAD. Thus, k = 2.5 for the normal distribution becomes c = 2.5 ⋅ 1.4826 = 3.7065 for this non-parametric version, and k = 3.0 for the normal distribution becomes c = 3.0 ⋅ 1.4826 = 4.4478."