(a) Single Proportion

1. Set up H_{0} : P = p_{0}.

2. Set up H_{1} : P > P_{0} or P < P_{0 }or P ≠ P_{0}·

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 Z_{tab}·

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. H_{0}: P = 116

2. H_{1}: 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, p_{0} = 1/6 = 0.167

= 1.98

6. Decision: Since Z_{cal}> 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. H_{0 }: P = 0.95 (components conforming specifications)

2. H_{1} : p < 0.95

4. α = 0.01, alternative hypothesis shows left tail test. Critical value is Z_{tab} = -2.33.

5. Computation :

Observed proportions = X= 185/220 = 0.84

6. Decision: Since Z_{cal }< -2 .33

=> Z_{cal} is inside the critical region.

=> H_{0} is rejected.

=> Manufacturer's claim cannot be accepted.

(b) Difference Two Proportions. Let p_{1 }and p_{2} be the proportions in two large samples of sizes n_{1} and n_{2} drawn respectively from two populations. To test whether the differences p_{1} - p_{2} as observed in the samples has arises only due to fluctuation of sampling.

1. Set up H_{0} : P_{1} = P_{2}

2. Set up H_{1 }: P_{1} =1= P_{2}

3. Test statistic

q = 1- p

Here, Z approximately follows standard normal distribution.

4. Set the level of significance and the critical value say, Z_{tab} using normal table.

5. Compute the statistic as Z_{cal}·

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 H_{0} : P_{1} = P_{2} (i.e., proportions of defectives before and after overhauling are equal).

2. H_{1}: P1 > P2

3. Test statistic:

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

Z_{tab} = 1.645.

5. Here n_{1} = 400, n2 = 300

P_{1} = 20/400 = 1/20, p_{2} = 10/300 = 1/30

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

6. Decision:

Since Z_{cal} < Z_{tab}

=> It lies in the acceptance region.

=> H_{0} is accepted.

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

=> Machine has not improved.