Hypothesis Testing about the Equality of Several Proportions Lecture 36

 

Karl Pearson’s Approximation

Lecture 36

Karl approximation transforms the discrete multinomial distribution so that it approaches a chi square distribution as n approaches infinity, establishing a connection between the discrete multinomial distribution and the chi square distribution. This approximation is widely used to test the agreement between observed and hypothesised data.

Suppose a random sample of n observations is distributed over k mutually exclusive and exhaustive cells. Let pi be the probability that an observation falls in the ith cell and ni be the number of observations falling in that cell such that ∑pi = 1 and ∑ni = n; that is, we have the following multinomial structure:

Cell	1	2	…..	i	….	K
Probability	P1	P2	…..	Pi	….	Pk	1
Frequency	n1	n2	…..	ni	….	nk	n

$$

Then the probability function follow multinomial distribution by

Testing hypothesis about P’s of the multinomial distribution

i. State the null and alternative hypotheses:

H0: Pi = Pi0 vs. H1: Pi ≠ Pi0     (i=1,2, ..., k)

ii. The significance level: α

iii. The test statistic: 

iv. Critical Region:

Reject H0 if  χ2 ≥ χ2 α (k-1)

v. Computation:

vi> Remarks.

Example 9.12: A machine is supposed to mix four dry fruits A, B, C, and D in the ratio of 3: 3: 2: 1. A can containing 500 of these mixed four dry fruits (170, 150, 100, and 80), respectively. Test the null hypothesis by the machine mixing the four fruits in the given ratio at the 5% significance level.

Solution: The total of 3, 3, 2, 1 is 9.

i. State the null hypotheses:

H0: P1 = 3/9, P2 = 3/ 9, P3 = 2/9, P4 = 1/9

H1: P1 ≠ 3/9, P2 ≠ 3/ 9, P3 ≠ 2/9, P4 ≠ 1/9

ii. The significance level: α = 0.05

iii. The test statistic: 

iv. Reject H0 if  χ2 ≥ χ2 0.05(3) = 7.815

v. Computation:

 nP10 = 500 × 3/9 = 166.67

 nP20 = 500 × 3/9 = 166.67

 nP30 = 500 × 2/9 = 111.11

 nP40 = 500 × 3/9 = 55.56

vi. Remarks: The chi square calculated value falls in the rejection region; the sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded the machine is not mixing the dry fruits in the mentioned ratio.

Example 9.12: The distribution of the numbers 0–9 in a random sample of 250 digits was as follows:

.

Digit	0	1	2	3	4	5	6	7	8	9
Frequency	17	31	29	18	14	20	35	30	20	36

Examine the hypothesis that the table's digits were distributed equally at the 0.05 level of significance.

Solution:

i. State the null and alternative hypotheses:

H0: Pi = 1/10 vs. H1: Pi ≠ 1/10     (i=0, 1, 2, ...,,9 )

ii. The significance level: α = 0.05

iii. The test statistic: 

iv. Reject H0 if  χ2 ≥ χ2 0.05(9) = 16.919

v. Computation:

nPi0 = 250 × 1/10 = 25

vi. Remarks: The chi square calculated value falls in the rejection region; the sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that the digits are not equally distributed.

Example 9.13: The local hospital provided the following records on the number of male and female births among 1000 families with four children:

No. of Births	0	1	2	3	4
Frequency	190	220	260	180	150

Test the null hypothesis that the male and female have the same chances.

Solution: The proportion of males is represented by P1, and the proportion of female birth is P2.

i. State the null and alternative hypotheses:

H0: Pi = 1/5 vs. H1: Pi ≠ 1/5     (i=0, 1, 2, 3, 4)

ii. The significance level: α = 0.05

iii. The test statistic: 

iv. Reject H0 if  χ2 ≥ χ2 0.05(4) = 9.488

v. Computation:

nPi0 = 800 × 1/5 = 160 

vi. Remarks: The chi square calculated value falls in the rejection region; the sample data does not provide sufficient evidence to accept the null hypothesis. Thus, it is concluded that the male and female births are not equally distributed in the five families.

Example 9.13: The number of heads appears 175 times after a coin is tossed 360 times. Test the hypothesis that coins are fair by using Karl Pearson and normal approximations.

Solution: The number of heads that appear is 175 and the number of tails that appear is 185.

Using Karl Approximation:

Let P1 be the proportion of head and P2 be the proportion of tail.

i. State the null and alternative hypotheses:

H0: Pi = 1/2 vs. H1: Pi ≠ 1/2     (i= H, T)

ii. The significance level: α = 0.05

iii. The test statistic: 

iv. Reject H0 if  χ2 ≥ χ2 0.05(1) = 3.841

v, Computation:

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 coin is fair.

Normal Approximation

Let P be the proportion of head 

i. State the null and alternative hypotheses:

H0: P = 1/2 vs. H1: P ≠ 1/2     

ii. The significance level: α = 0.05

iii. The test statistic: 

vi. Reject H0 when |z|≥ 1.96

v. Computation:

np0 = 360  × 1/2 = 180

x < np0

vi. Remarks: The z 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 coin is fair.

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