Answer:
H₀ should be rejected at CI 95% .
Then the proportion of adults with 95 % CI is bigger than the proportion of teen
Step-by-step explanation:
From adult sample:
n₂ = 2323
x₂ = 1601
p₂ = 1601 / 2323 p₂ = 0,689 or p₂ = 68,9%
From teen sample:
n₁ = 875
x₁ = 555
p₁ = 555/ 875 p₁ = 0,634 or p₁ = 63,4
Values of p₁ and p₂ suggest that the proportion of adults consuming at least one meat free meal per week is bigger than teen proportion.
To either prove or reject the above statement we have to develop a difference of proportion test according to:
Hypothesis test:
Null Hypothesis H₀ p₁ = p₂
Alternative Hypothesis Hₐ p₂ > p₁
So is a one-tail test to the right
We can establish a confidence interval of 95 % then α = 5 %
or α = 0,05
As the samples are big enough we will develop a z test
Then z(c) for α = 0,05 from z table is z(c) = 1,64
To calculate z(s)
z(s) = ( p₂ - p₁ ) / √p*q* ( 1/n₁ + 1/n₂ )
where p = ( x₁ + x₂ ) / n₁ + n₂ p = 555 + 1601 / 875 + 2323
p = 2156/3198 p = 0,674
and q = 1 - 0,674 q = 0,326
z(s) = ( 0,689 - 0,634 ) / √0,674*0,326 ( 1/875 + 1 / 2323
z(s) = 0,055/ √ 0,2197 ( 0,00114 + 0,00043)
z(s) = 0,055/ √0,2197* 0,00157
z(s) = 0,055/ √ 3,45*10⁻⁴
z(s) = 0,055 / 1,85*10⁻²
z(s) = 5,5/1,85
z(s) = 2,97
Comparing z(s) and z(c) z(s) > z(c)
Then z(s) is in the rejection region we reject H₀.
We can claim that the proportion of adult eating at least one meat-free meal is bigger than the proportion of teen
