Respuesta :
Answer:
[tex]z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318[/tex]
[tex]p_v =P(Z>0.318)=0.375[/tex]
So the p value is a very high value and using the significance level provided [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] 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
[tex]X_{1}=507[/tex] represent the number of children diagnosed with Autism Spectrum Disorder (ASD) in Arizona
[tex]n_{1}=32601[/tex] sample elected from arizona
[tex]\hat p_{1}=\frac{507}{32601} =0.0156[/tex] represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona
[tex]p_{2}=\frac{1}{88}= 0.0114[/tex] represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in the nation
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]p_{1} \leq p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} > p_{2}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{507+1}{32601 +88}=0.01554[/tex]
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318[/tex]
Statistical decision
The significance level provided is [tex]\alpha=0.01[/tex], and we can calculate the p value for this test.
Since is a one right sided test the p value would be:
[tex]p_v =P(Z>0.318)=0.375[/tex]
So the p value is a very high value and using the significance level provided [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] 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.
Testing the hypothesis, it is found that since the p-value of the test is of 0 < 0.01, there is sufficient data to show that the incident of ASD is more in Arizona than nationally.
At the null hypothesis, it is tested if the proportion in Arizona is the same as the national proportion, that is:
[tex]H_0: \mu = \frac{1}{88} = 0.0114[/tex]
At the alternative hypothesis, it is tested if the proportion is greater, that is:
[tex]H_1: \mu > 0.0114[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are: [tex]p = 0.0114, n = 32601, \overline{p} = \frac{507}{32601} = 0.0156[/tex]
The value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.0156 - 0.0114}{\sqrt{\frac{0.0114(0.9886)}{32601}}}[/tex]
[tex]z = 7.14[/tex]
The p-value of the test is the probability of a finding a sample proportion above 0.0156, which is 1 subtracted by the p-value of z = 7.14.
z = 7.14 has a p-value of 1.
1 - 1 = 0.
Since the p-value of the test is of 0 < 0.01, there is sufficient data to show that the incident of ASD is more in Arizona than nationally.
A similar problem is given at https://brainly.com/question/24166849
