P(-15)
s(-2)
=x(x-5)+3(x-5)
=(x+3)(x-5)
x=5
1 answer:
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P(-15)
s(-2)
=x(x-5)+3(x-5)
=(x+3)(x-5)
x=5
Answer:
So the p value is a very high value and using the significance level provided we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona is not significantly higher than the national incident rate at 1% of significance.
Step-by-step explanation:
Data given and notation
represent the number of children diagnosed with Autism Spectrum Disorder (ASD) in Arizona
sample elected from arizona
represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona
represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in the nation
z would represent the statistic (variable of interest)
represent the value for the test (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to check if that the incident of ASD is more in Arizona than nationally, the system of hypothesis are:
Null hypothesis:
Alternative hypothesis:
We need to apply a z test to compare proportions, and the statistic is given by:
(1)
Where
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
Statistical decision
The significance level provided is , and we can calculate the p value for this test.
Since is a one right sided test the p value would be:
So the p value is a very high value and using the significance level provided we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona is not significantly higher than the national incident rate at 1% of significance.