Answer:
p-value: 0.6527
Explanation:
Hello!
You have two samples to study, from each sample the weight of each child was measured and counted the total of overweight kids (x: "success") in each group:
Sample 1 (Boys aged 6-11)
n₁= 546
x₁= 89
^p₁= x₁/n₁ = 89/546 ≅0.16
Sample 2 (girls aged 6-11)
n=508
x= 74
^p= x/n = 74/508 ≅ 0.15
If the hypothesis statement is "The proportion of boys that are overweight differs from the proportion of girls that are overweight", the test hypothesis is:
H₀: ρ₁ = ρ₂
H₁: ρ₁ ≠ ρ₂
This type of hypothesis leads to a two-tailed rejection region, then the p-value will also be two-tailed. To calculate the p-value you have to first calculate the value of the statistic under the null hypothesis, in this case, is a test for the difference between two proportions:
Z= (^ρ₁ - ^ρ₂) - (ρ₁ - ρ₂) ≈ N(0;1)
√(ρ` * (1 - ρ`) * (1/n₁ + 1/n₂))
ρ`= x₁ + x₂ = 89+74 = 0.154 ≅ 0.15
n₁ + n₂ 546 + 508
Z⁰ᵇ = (0.16-0.15) - (0)
√(0.15 * (1 - 0.15) * (1/546 + 1/508))
Z⁰ᵇ = 0.45
I've mentioned before that in this test you have a two-tailed p-value. The value calculated (0.45) corresponds to the right or positive tail and the left tail is symmetrical to it concerning the distribution mean, in this case, is 0, so it is -0.45. To obtain the p-value you need to calculate the probability of both values and add them:
P(Z>0.45) + P(Z<-0.45) = (1- P(Z<0.45)) + P(Z<-0.45) = (1-0.67364) + 0.32636 = 0.65272 ≅ 0.6527
p-value: 0.6527
Since there is no signification level in the problem, I'll use the most common to reach a decision. α: 0.05
Since the p-value is greater than α, you do not reject the null Hypothesis, in other words, there is no significative difference between the proportion of overweight boys and the proportion of overweight girls.
I hope it helps!