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fenix001 [56]
3 years ago
13

A study was performed to investigate whether teens and adults had different habits when it comes to consuming meat-free meals. I

n particular, the researchers were interested in the relationship between p1, the proportion of teens who would report eating at least one meat-free meal in the past week, and p2, the proportion of adults who would report eating at least one meat-free meal in the past week. A random sample of 875 teens and a separate random sample of 2,323 adults found that 555 of the teens and 1,601 of the adults reported eating at least one meat-free meal in the past week. The conditions for inference were checked and verified. What is the correct set of hypotheses the researchers should test in this scenario
Mathematics
1 answer:
Elden [556K]3 years ago
6 0

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

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