Question

Twenty First Century Laundry
69	3	1	72	93	3	80	6	5	36
42	2	64	44	15	77	49	85	82	102
28	70	41	30	25	45	31	49	49	49
13	42	27	13	45	31	14	34	32	39
19	29	12	19	30	14	21	14	14	52
7	13	18	8	14	19	18	21	21	5
94	19	6	107	19	2	41	7	15	15
53	8	90	55	2	64	16	12	27	53
35	16	35	38	3	9	9	9	11	39
16	17	18	23	74	25	26	58	59	62

Construct and show one box plot for the overall data. clearly show the values for the 5 box plot statistics

What are the values for the upper outlier and lower outlier limits?

Are there any outliers?

Answer 1

Solution:

We have the following summary statistics for the given data on Twenty-First-Century Laundry:

Case Processing Summary
	Cases
	Valid		Missing		Total
	N	Percent	N	Percent	N	Percent
Tweenty_First_Century_Laundry	100	100.0%	0	0.0%	100	100.0%

Descriptive Statistics
	N	Minimum	Maximum	Mean	Std. Deviation
Tweenty_First_Century_Laundry	100	1	107	32.92	25.899
Valid N (listwise)	100

Descriptives
			Statistic	Std. Error
Tweenty_First_Century_Laundry	Mean		32.92	2.590
	95% Confidence Interval for Mean	Lower Bound	27.78
		Upper Bound	38.06
	5% Trimmed Mean		31.07
	Median		25.50
	Variance		670.741
	Std. Deviation		25.899
	Minimum		1
	Maximum		107
	Range		106
	Interquartile Range		35
	Skewness		.994	.241
	Kurtosis		.233	.478

The box plot is given below:

Statistics
Tweenty_First_Century_Laundry
N	Valid	100
	Missing	0
Percentiles	25	14.00
	50	25.50
	75	49.00

We see that first quartile Q1 = 14 and third quartile Q3 = 49.

So, Inter Quartile Range(IQR) = (Q3-Q1) = ( 49 - 14 ) = 35.

So, the lower outlier limit, L = Q1 - 1.5*IQR = 14 - 1.5*35 = -38.5.

Since our all observations are positive, L = 0.

The upper outlier limit is U = Q3+1.5*IQR = 49 + 1.5*35 = 101.5.

From the box plot, we observe that there two outliers and they are given below: 107 and 102.