The Median Test Lecture 53

 The Median Test 

Lecture 53

The median test is a non-parametric version of the one-sample and two-sample t-tests and one-way ANOVA. The median test is used to test the null hypothesis that the sample(s) drawn from the population(s) have the same median. The alternative hypothesis can be either that the median is equal to a specified value or two or more medians are different (two-tailed test) or that one median is greater than the other (one-tailed test). The median test is based on the following assumptions.

The observations in the sample(s) can be measured in terms of an ordinal scale.

Each observation in the sample(s) occurs in the random manner.

Each observation is independent of the other observations.

The two or more samples will be independent.

One Sample Median Test

The one-sample median test is used to test the hypothesis that the median of a sample is equal to the population hypothesised median.

Procedure: Place the number of observations above and below the hypothesised value of the median in the following 2 X 2 table and find the expected frequency as:

Expected Frequency = n / 2

Where:

n is the number of observations.

	Observed Frequency	Expected Frequency	Total
Above median	a	b	(a+b)
Below median	c	d	(c+d)
Total	(a+c)	(b+d)	n

The test statistic to be used is

 Median test in case of two samples

Find the median of the combined observations of both samples and develop a table

		Sample 2	Total
Above Median	a	b	(a + b)
Below median	c	d	(c + d)
Total	(a + c)	(b + d)	n

The test statistic is to be used is

Median test in case of k samples

Given k samples with n₁, n₂, ...,and nₖ observations, compute the grand median of all n₁ + n₂ + ... + nₖ observations. Then construct a 2xk contingency table as:

The test statistic is to be used is

 Follow chi-square with (k – 1) degrees of freedom.

Example 13.4: The policy of an administration those students are eligible whose median is at least 65. The scores of a student in the last five examinations are 64, 67, 75, 80, 56, and 70. Test the null hypothesis: the student is eligible for admission.

Solution:

i. State the null and alternative hypotheses

H0: Median ≥ 65 vs. H1: Median < 65

ii. The significance level; α = 0.05

iii. The test statistic:

iv. Reject H0 if ${\chi }^2 \le {{\chi }^{\mathrm{2 }}}_{0.95} \left(1\right) =0.004$

$\mathrm{v.\; Computation:\; The\; observed\; frequency\; below\; and\; above\; the\; median\; is\; 2\; and\; 4,\; respectively,\; and\; expected\; frequencies\; can\; be\; obtained\; as:}$

$\mathrm{Expected\; Frequency\; =\; n\; /\; 2 }$

$\mathrm{Expected\; Frequency\; =\; 6\; / }$2

 Expected Frequency = 3

	Observed Frequency	Expected Frequency	Total
Above Median	4	3	7
Below median	2	3	5
Total	6	6	12

vi. Remarks: The chi square calculated value falls in the acceptance region; the sample data does not provide sufficient evidence to reject the null hypothesis. Thus, it is concluded that the student is eligible for admission.

Example 13.5: The following data is obtained from patients recovered from scabies treated by two different techniques of the same medicine.

Method 1	92	63	30	78	24	19	26	79	54	57
	97	46	58	74	77	80	93	99	78	50
Method 2	77	87	99	62	76	47	66	83	72	80
	53	80	48	75	76	78	97	53	64	67

Use the median test at the 5% level to test the null hypothesis that the two samples are drawn from a population with the same median of recovery.

Solution:

i. State the null and alternative hypotheses:

H0: Median 1 = Median 2 vs. H1: Median 1 ≠ Median 2 

ii. The significance level; α = 0.05

iii. The test statistic: 

iv. Reject H0, if  χ2 ≥ χ20.05(1) = 3.84

v. Computation: Find the median of the combined observations of both samples.

The combined median is 75.50

	Sample 1	Sample 2	Total
Above Median	10	10	20
Below median	10	10	20
Total	20	20	40

vi. Remark: The chi-square calculated value falls in the acceptance region. The sample data does not provide sufficient evidence to reject the null hypothesis. Thus, we conclude that the medians of recovery for both populations from which samples are selected have identical medians.

Example 13.6: Four random samples, each of size eight, are selected from four populations and are given below:

Sample 1	29	40	43	30	46	34	32	27
Sample 2	35	39	33	31	40	34	33	29
Sample 3	34	43	42	31	35	40	35	42
Sample 4	33	34	32	26	48	30	35	42

Use the median test to test the hypothesis that all four samples come from an identical population.

Solution: 

i. State the null and alternative hypotheses:

H0: Median 1 = Median 2 = Median 3 = Median 4 

vs. 

H1: Median 1 ≠ Median 2 ≠ Median 3 ≠ Median 4 

ii. The significance level; α = 0.05

iii. The test statistic:

iv. If χ2 ≥ χ20.05(3) = 7.82, reject H0.

v. Computation: find the median of the combined observations of all samples.

The median = 34

Drop the observation that is equal to the median and construct the following table:

	Sample 1	Sample 2	Sample 3	Sample 4	Total
Above Median	3	3	5	3	14
Below Median	4	4	2	4	14
Total	7	7	7	7	28

vi. Remarks: The chi-square calculated value falls in the acceptance region. The sample data does not provide sufficient evidence to reject the null hypothesis. Thus, we conclude that the medians of the four populations are identical.

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