Question

A June 2009 Harris Interactive poll asked people their opinions about the influence of advertising on the products they buy. Among the people aged 18 to 34 years, 45% view advertisements as being influential, whereas among the people aged 35 to 44 years, 37% view advertisements as being influential. Suppose that this survey included 655 people in the 18- to 34-year age group and 420 in the 35- to 44- year age group. At the 1% significance level, can you conclude that the proportion of all people aged 18 to 34 years who view advertisements as being influential in their purchases is greater than the proportion of all people aged 35 to 44 years who hold the same opinion

Answer 1

We accept with 99 % of confidence that p₁ ( the proportion in the group aged 18 - 34 ) is bigger than p₂ (the proportion in the group aged 35-44)

Explanation:

Group aged 18 - 34

Sample size n₁ = 655 sample size enough to apply a normal distribution aproximation for the test

p₁ = 0,45 and q₁ = 1 - p₁ q₁ = 0,55

p₁*n₁ = 0,45*655 = 294 q₁*n₁ = 0,55*655 = 360

both bigger than 5

p₁ = 45 % or p₁ = 0,45 then q₁ = 1 - 0,45 q₁ = 0,55

Group aged 35 - 44

Sample size n₂ = 420 sample size enough to apply a normal distribution aproximation for the test

p₂ = 37% p₂ = 0,37 and q₂ = 0,63

p₂*n₂ = 0, 37*420 = 155 q₁*n₁ = 0,63*420 = 264

Test Hypothesis

Null Hypothesis H₀ p₁ = p₂

Alternative Hypothesis Hₐ p₁ > p₂

Significance level α = 1% α = 0,01

Confidence Interval CI = 99 %

Then from z table we find critical z z (c ) = 3,1

Calculating z(s)

z(s) = ( p₁ - p₂ )/ √[ ( p₁*q₁)/n₁ ] + [ (p₂*q₂)/n₂ ]

z(s ) = (0,45 - 0,37 ) / √ 0,45*0,55)/655 + 0,37*0,63 /420

z(s) = 0,08 /0,03

z(s) = 2,66

Comparing z(s) and z(c)

z(s) < z(c)

Then z (s) is inside the acceptance region for H₀