Question

The below age variable was inputted into SPSS and the descriptive statistics output generated. I did the interquartile range (39) to try and answer Question #4: Are there outliers among the values of age? provide a rationale for your answer. need help determining the Q1, Q3 if this is the correct approach to answer this questions and respond to what's the rationale?

Age variable
42
41
56
78
86
49
82
35
59
37

Descriptives
			Statistic	Std. Error
Age	Mean		56.50	6.091
	95% Confidence Interval for Mean	Lower Bound	42.72
		Upper Bound	70.28
	5% Trimmed Mean		56.06
	Median		52.50
	Variance		370.944
	Std. Deviation		19.260
	Minimum		35
	Maximum		86
	Range		51
	Interquartile Range		39
	Skewness		.538	.687
	Kurtosis		-1.393	1.334

Answer 1

To determine outliers we have to calculate Q1 and Q3.

So first of all we have to write the data in increasing order i.e,

35,37,41,42,49,56,59,78,82,86

Now for Q1 is the data point above which 25 percent of the data sits.

Q1=(N+1)/4 th term=11/4=2.75th term=2nd term+0.75*(3rd-2nd)=37+3=40

And Q3 is the data point above which 75 percent of the data sits.So

Q3=3*(N+1)/4 th term=33/4=8.25th term= 8th term+0.25*(9th - 8th)=78+1=79

So interquatile range=Q=Q3-Q1=79-40=39 so it is as given in here.

Outliers are identified by assessing whether or not they fall within a set of numerical boundaries called "inner fences" and "outer fences".A point that falls outside the data set's inner fences is classified as a minor outlier, while one that falls outside the outer fences is classified as a major outlier. To find the inner fences for your data set, first, multiply the interquartile range by 1.5. Then, add the result to Q3 and subtract it from Q1. The two resulting values are the boundaries of your data set's inner fences.

Now for inner fences of our data are Q3+( 39*1.5=58.5)=79+58.5=137.5

And Q1-58.5=40-58.5= -18.5

So the boundaries are -18.5 and 137.5

So no data lies out side inner fences.We don't have any minor outliers as well as any major outliers.