Mathematics, 02.03.2020 21:22 saltyclamp

Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated ("Autism and developmental," 2008). Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah? Test at the 1% level.

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Answer from: Jasoncookies23

$z=\frac{0.0211-0.0132}{\sqrt{0.0141(1-0.0141)(\frac{1}{18440}+\frac{1}{2123})}}=24.827$

$p_v =P(Z24.827)\approx 0$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of children with ASD in Utah is significantly higher than the proportion of children with ASD in Pennsylvania

Step-by-step explanation:

Data given and notation

$X_{P}=245$ represent the number of children with ASD in Pennsylvania

$X_{U}=45$ represent the number of children with ASD in Utah

$n_{P}=18440$ sample for Pennsylvania

$n_{U}=2123$ sample for Utah

$p_{P}=\frac{245}{18440}=0.0133$ represent the proportion of children with ASD in Pennsylvania

$p_{U}=\frac{45}{2123}=0.0212$ represent the proportion of children with ASD in Utah

z would represent the statistic (variable of interest)

p_v 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 the proportion for Utah is higher than the proportion for Penssylvania , the system of hypothesis would be:

Null hypothesis: $p_{U} \leq p_{P}$

Alternative hypothesis: $p_{U} \mu_{p}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{U}-p_{P}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{U}}+\frac{1}{n_{P}})}}$ (1)

Where $\hat p=\frac{X_{U}+X_{P}}{n_{U}+n_{P}}=\frac{245+45}{18440+2123}=0.0141$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.0211-0.0132}{\sqrt{0.0141(1-0.0141)(\frac{1}{18440}+\frac{1}{2123})}}=24.827$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one right tailed test the p value would be:

$p_v =P(Z24.827)\approx 0$

Answer from: Quest

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to $\frac{1}{60} hr\\$

then mult. both sides by 75 we get the ratio

75 mi. to $\frac{75}{60} hr\\$

simplifying we get

75 mi. to $\frac{5}{4} hr$

or 1.25 hr.

Answer from: Quest

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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