Assume that last year in a particular state with 700000 children there were 1205 children out of 32600 in a random sample who were diagnosed with Autism Spectrum Disorder. Nationally, 1 out of 88 children are diagnosed with ASD. Is there sufficient data to show that the incident rate of ASD is higher in that state than nationally

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

whatare you sayig

step-by-step explanation:

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!