Answer:
We accept Hₐ there is enough evidence to claim that the incidence of ASD is bigger in this state than it is nationally
Step-by-step explanation:
The population follows a binomial distribution, and the fact that
p₀ = 1205/32600 = 0,03696 and then q₀ = 1-0,03696 = 0,9630
And both
p₀*n = 0,03696*32600 = 1205 > 10
q₀*n = 31395 > 10
The binomial distribution can be approximate to a Normal Distribution, then we need to develop a one tail test ( to the right) to evaluate if the claim " There is sufficient data to show that the incident rate is higher in the mentioned state than the national one.
Therefore:
Hypothesis Test H₀ ⇒ p₀ = p
Alternative Hypothesis Hₐ ⇒ p₀ > p
p is the population proportion mean
p = 1/88 = 0,011
We must test having a very wide Confidence Interval since the matter we are trying is public health then we choose CI = 95 %
CI = 95 % α = 5 % α = 0,05
Critical value for 0,05 is from z-table z(c) = 1,64
We compute z(s)
z(s) =( p₀ - p )/ √ p₀*q₀/n
z(s) = 0,03696 - 0,011 / √ (0,03696*0,9630)/ 32600
z(s) = 0,02596* 180,55/ √ 0,0356
z(s) = 4,687 / 0,189
z(s) = 24,9
Comparing z(s) and z(c) 24,9 > 1, 64
z(s) is out of the acceptance region and we have to reject H₀
We accept Hₐ there is enough evidence to claim that the incidence of ASD is bigger in this state than it is nationally
