Hypothesis Testing Of The Difference Between Proportions
Illustration
Ken industrial producer have manufacture a perfume termed as "fianchetto." In order to test its popularity or reputation in the market, the manufacturer carried a random surveyor study in back rank city whereas 10,000 consumers were interviewed after that 7,200 demonstrated preferences. The manufacturer moved also to area Rook town where he interviewed 12,000 consumers out of that 1,0000 demonstrated preference for the product.
Required
Design a statistical test and thus use it to advise the manufacturer regarding the differences in the proportion, at 5 percent level of significance.
Solution
H_{0} : π_{1} = π_{2}
H_{1} : π_{1} ≠ π_{2}
The critical value for this two tailed test at 5 percent level of significance = 1.96.
Now Z = ¦{(P_{1}  P_{2})  (Π_{1}  Π_{2})/S(P_{1}  P_{2})}¦
But as the null hypothesis is π_{1} = π_{2}, the second part of the numerator disappear that is
π_{1}  π_{2} = 0 which will usually be the case at this level.
Then Z = ¦ {(P_{1}  P_{2})/S (P_{1}  P_{2})}¦
Whereas:

Sample 1

Sample 2

Sample size

n_{1} = 10,000

n_{2} = 12,000

Sample proportion of success

P_{1} =0.72

P_{2} = 0.83

Population proportion of success.

Π_{1}

Π_{2}

Here S (P_{1}  P_{2}) = √{(pq/n_{1}) + √(pq/n_{1})}
Whereas p = (p_{1}n_{1} + p_{2}n_{2})/ (n_{1} + n_{2})
And q = 1  p;
∴ in our case
P = {10,000 (0.72) + 12,000 (0.83)}/(10,000 + 12,000)
= 84,000/22,000
= 0.78
∴ q = 0.22
S (P_{1}  P_{2}) = √{(0.78(0.22)/10,000) + (0.78(12,000)/12,000)}
= 0.00894
= ¦(0.72  0.83)/0.00894¦
=12.3
As 12.3 > 1.96, we reject the null hypothesis however accept the alternative. The differences among the proportions are statistically significant. It implies that the perfume is more popular in Rook town rather than in Back rank city.