1. The document discusses hypothesis testing for single and double proportions using examples. 2. For a single proportion example, it tests whether there is increased interest in extra spicy ketchup compared to 2 years ago. It finds the current interest is significantly greater. 3. For a double proportion example, it tests whether brand A outsells brand B for cigarettes. It finds brand A outsells brand B. 4. A third example tests a cigarette company's claim that brand A outsells brand B by 8%. It finds the 8% difference claim is valid.

1. HYPOTHESIS TESTING
PART-V
SINGLE PROPORTION
DOUBLE PROPORTION
NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS

2. Test Concerning Single Proportion
Example-1
A ketchup manufacturer is in the process of deciding whether to
produce a new extra-spicy brand. The company’s marketing-
research department used a national telephone survey of 6000
households and found that the extra-spicy ketchup would be
purchased by 335 of them. A much more extensive study made 2
years ago showed that 5 percent of the households would
purchased the brand then. At a 2 percent significance level,
should the company conclude that there is an increased interest in
the extra-spicy flavor?

3. Solution:
n = 6000, 𝑥 = 335
𝑝 =
𝑥
𝑛
=
335
6000
= 0.05583
P = 0.05, Q = 0.95, α = 0.02
1.Hypothesis
H0: P = 0.05
H1: P > 0.05
2.Level of significance α = 0.02
3.Test statistic
ˆp P
z
PQ
n



4. 4. Critical Region
5.Computation 0.05583 0.05
0.05(0.95)
6000
2.07
Z
Z
cal



6.Conclusion: Reject H0
(The current interest is significantly greater than
the interest of 2 years ago).
 –0– 2.05
In case of upper tail test i.e. H1 𝑖𝑠 ˃.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 = 𝑍0.02 = 2.05
𝑍 𝑐𝑎𝑙 ≥ 2.05
(Using inverse area of normal table)

5. Test Concerning Double Proportion
Example-2
A cigarette manufacturing firm distributes two brands of
cigarettes, if it is found that 56 to 200 smokers prefer
brand A and that 29 of 150 smokers prefer brand B, can
we conclude at the 0.06 level of significance that brand
A outsells brand B?

6. Solution:
𝑛 𝐴 = 200 𝑛 𝐵 = 150
𝑥 𝐴 = 56 𝑥 𝐵 = 29
𝑝 𝐴 =
𝑥 𝐴
𝑛 𝐴
=
56
200
= 0.28 𝑝 𝐵 =
𝑥 𝐵
𝑛 𝐵
=
29
150
= 0.193
𝑃 =
𝑥 𝐴+𝑥 𝐵
𝑛 𝐴+𝑛 𝐵
=
56+29
200+150
=
85
350
= 0.24
𝑄 = 1 − 𝑃 = 1 − 0.24 = 0.76
α = 0.06
1.Hypothesis
H0: 𝑃𝐴 = 𝑃𝐵
H1: 𝑃𝐴 > 𝑃𝐵
2.Level of significance α = 0.06

7. 3.Test statistic z =
pA−pB
𝑃 𝑄(
1
nA
+
1
nB
)
4.Critical Region
In case of upper tail test i.e. H1 𝑖𝑠 ˃.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼 = 𝑍0.06 = 1.56
𝑍 𝑐𝑎𝑙 ≥ 1.56 (Using inverse area of normal table)
 –0– 1.56
5.Computation
z =
0.28 − 0.193
(0.24)(0.76)(
1
200
+
1
150
)

8. 6.Conclusion: Reject H0(Brand A outsells brand B).
z =
0.087
(0.1824)(0.005 + 0.0066)
z =
0.087
0.00211
z =
0.087
0.046
z = 1.89

9. Example-3
A cigarette manufacturing firm claims that its brand A line of
cigarettes outsells its brand B line by 8%. If it is found that 40 of
200 smokers prefer brand A and 15 of 150 smokers prefer brand
B, test whether the 8% difference is a valid claim at 1% level of
significance.
Solution:
𝑛 𝐴 = 200 𝑛 𝐵 = 150
𝑥 𝐴 = 40 𝑥 𝐵 = 15
𝑝 𝐴 =
𝑥 𝐴
𝑛 𝐴
=
40
200
= 0.20 𝑝 𝐵 =
𝑥 𝐵
𝑛 𝐵
=
15
150
= 0.10
𝑞 𝐴 = 1 − 𝑝 𝐴 = 1 − 0.20 = 0.80
𝑞 𝐵 = 1 − 𝑝 𝐵 = 1 − 0.10 = 0.90

10. α = 0.01
1.Hypothesis
H0: 𝑃𝐴 - 𝑃𝐵 = 0.08
H1: 𝑃𝐴 - 𝑃𝐵 ≠ 0.08
2.Level of significance α = 0.01
3.Test statistic 𝑧 =
𝑝 𝐴− 𝑝 𝐵 − 𝑃 𝐴−𝑃 𝐵
𝑝 𝐴 𝑞 𝐴
𝑛 𝐴
+
𝑝 𝐵 𝑞 𝐵
𝑛 𝐵

11. 4.Critical Region
In case of two tail test i.e. H1 𝑖𝑠 ≠.
Reject H0, if 𝑍 𝑐𝑎𝑙 ≤ −𝑍𝑡𝑎𝑏 or 𝑍 𝑐𝑎𝑙 ≥ 𝑍𝑡𝑎𝑏.
Where 𝑍𝑡𝑎𝑏 = 𝑍 𝛼
2
= 𝑍0.01
2
= 𝑍0.005 = 2.58
𝑍 𝑐𝑎𝑙 ≤ −2.58 or 𝑍 𝑐𝑎𝑙 ≥ 2.58.
(Using inverse area of normal table)

12. 5.Computation
𝑧 =
𝑝 𝐴 − 𝑝 𝐵 − 𝑃𝐴 − 𝑃𝐵
𝑝 𝐴 𝑞 𝐴
𝑛 𝐴
+
𝑝 𝐵 𝑞 𝐵
𝑛 𝐵
𝑧 =
0.20 − 0.10 − 0.08
(0.20)(0.80)
200
+
(0.10)(0.90)
150
𝑧 =
0.02
0.0008 + 0.0006

13. 𝑧 =
0.02
0.037
𝑧 = 0.54
6.Conclusion:
Accept H0(We may conclude that a difference of 8% in
the sale of two brands of cigarettes is a valid claim by
the firm).