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Testing of Proportion

(a) Single Proportion

1. Set up H0 : P = p0.

 

2. Set up H1 : P > P0 or P < P0 or P ≠ P0·

 

3. Set up the test statistics

which approximately follows the standard normal distribution.

 

4. Set up level of significance a and the critical value says Ztab·

 

5. Compute the statistic.

 

6. Decisions:

Here the conclusions are the similar as given in section 2(a) for testing of a single mean of a large sample.

 

Example 9. A die was thrown 500 times and six resulted 100 times. Do the data justify the hypothesis of an unbiased die ?

 

Solution. Let us assume that the die is unbiased and the probability of obtaining a six with the die is 116.

 

1. H0: P = 116

2. H1: P ≠ 116

3. Test statistic:

4. Let α = 0.05. Alternative hypothesis suggests for two tailed test.

α/2 = 0.025, critical values are -1.96 and 1.96.

5. Computation: Here out of 500 times throw, six resulted 100 times. So the observed value of proportion (X) of six is

 

X= 100/500 = 0.2, p0 = 1/6 = 0.167

= 1.98

 

 

6. Decision: Since Zcal> 1.96

=>                    H0 is rejected

=>                    The given data do not justify the hypothesis of an unbiased die.

 

Example 10. A manufacturer claimed that at least 95% of the components of an electronic Circuit board which he supplied conformed to specifications. A random sample of 220 components showed that only 185 were up to the standard. Test his claim at 1% level of significance.

Solution.         1. H0 : P = 0.95 (components conforming specifications)

2. H1 : p < 0.95

3. Test statistics:

4.  α = 0.01, alternative hypothesis shows left tail test. Critical value is Ztab = -2.33.

5. Computation :

Observed proportions = X= 185/220 = 0.84

n = 220

6. Decision: Since Zcal < -2 .33

=>                    Zcal is inside the critical region.

=>                    H0 is rejected.

=>                    Manufacturer's claim cannot be accepted.

 

(b) Difference Two Proportions. Let p1 and p2 be the proportions in two large samples of sizes n1 and n2 drawn respectively from two populations. To test whether the differences p1 - p2 as observed in the samples has arises only due to fluctuation of sampling.

 

1. Set up H0 : P1 = P2

2. Set up H1 : P1 =1= P2

3. Test statistic

q = 1- p

 

Here, Z approximately follows standard normal distribution.

 

4. Set the level of significance and the critical value say, Ztab using normal table.

5. Compute the statistic as Zcal·

6. Decision:

Here, the decisions are the similar as given in testing of difference of two means for large samples in section 2 (b).

 

Example 11. A machine produced 20 defective articles in a batch of 400. After overhauling it produced 10 defectives in a batch of 300. Has the machine improved?

 

Solution. 1. Set up H0 : P1 = P2 (i.e., proportions of defectives before and after overhauling are equal).

2. H1: P1 > P2

3. Test statistic:

4. Let  α = 0.05. Alternative hypothesis shows it is a right tailed test. So the critical value is

Ztab = 1.645.

5. Here                        n1 = 400, n2 = 300

P1 = 20/400 = 1/20,                             p2 = 10/300 = 1/30

 

P = (20+10)/(400+300) = 3/70,          q = 1 - p = 67/70

6. Decision:

Since Zcal < Ztab

 

=>                    It lies in the acceptance region.

=>                    H0 is accepted.

=>                    Two population proportions before and after overhauling are equal.

=>                    Machine has not improved.