Assume that last year in a particular state with 700000 children there were 1022 children out of 31800 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? Test at the 4% level.P: PARAMETER What is the correct parameter symbol for this problem? Correct What is the wording of the parameter in the context of this problem? CorrectH: HYPOTHESES Fill in the correct null and alternative hypotheses: The value of p on this one is a fraction....leave it exact. 𝐻0: Correct Correct 188Correct 𝐻𝐴: Correct Correct 188Correct A: ASSUMPTIONS Since Correct information was collected from each object, what conditions do we need to check? Check all that apply. 𝑛≥30 or normal population.𝑛(1-𝑝)≥10𝑛(𝑝̂)≥10σσ is known.𝑛𝑝≥10σσ is unknown.𝑁≥20𝑛𝑛(1-𝑝̂)≥10Correct Check those assumptions: Round computations to 2 decimal places. 1. 𝑛𝑝 = $31,8000$31,8000Incorrect which is Correct 10Correct 2. 𝑛(1-𝑝) = $31,8000$31,8000Incorrect which is Correct 10Correct 3. 𝑁 = 700,000Correct which is Correct 636,000Correct If no N is given in the problem, use 1000000N: NAME THE PROCEDURE The conditions are met to use a Correct .T: TEST STATISTIC The symbol and value of the random variable on this problem are as follows: Leave this answer as a fraction. Correct = 102231800Correct The formula set up of the test statistic is as follows.: (Leave any values that were given as fractions as fractions) 𝑧=𝑝̂-𝑝𝑝(1-𝑝)𝑛=( 102231800Correct -188Correct )/((188Correct ⋅(1-188Correct )) /31800Correct ) Final answer for the test statistic from technology. Round to 2 decimal places: z = 34.73IncorrectO: OBTAIN THE P-VALUE Report to 4 decimal places. It is possible when rounded that a p-value is 0.0000 P-value = .0000CorrectM: MAKE A DECISION Since the p-value Correct .04Correct , we Correct .S: STATE A CONCLUSION There Correct significant evidence to conclude Correct Correct
Question
Assume that last year in a particular state with 700000 children there were 1022 children out of 31800 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? Test at the 4% level.P: PARAMETER What is the correct parameter symbol for this problem? Correct What is the wording of the parameter in the context of this problem? CorrectH: HYPOTHESES Fill in the correct null and alternative hypotheses: The value of p on this one is a fraction....leave it exact. 𝐻0: Correct Correct 188Correct 𝐻𝐴: Correct Correct 188Correct A: ASSUMPTIONS Since Correct information was collected from each object, what conditions do we need to check? Check all that apply. 𝑛≥30 or normal population.𝑛(1-𝑝)≥10𝑛(𝑝̂)≥10σσ is known.𝑛𝑝≥10σσ is unknown.𝑁≥20𝑛𝑛(1-𝑝̂)≥10Correct Check those assumptions: Round computations to 2 decimal places. 1. 𝑛𝑝 = 31,8000Incorrect which is Correct 10Correct 2. 𝑛(1-𝑝) = 31,8000Incorrect which is Correct 10Correct 3. 𝑁 = 700,000Correct which is Correct 636,000Correct If no N is given in the problem, use 1000000N: NAME THE PROCEDURE The conditions are met to use a Correct .T: TEST STATISTIC The symbol and value of the random variable on this problem are as follows: Leave this answer as a fraction. Correct = 102231800Correct The formula set up of the test statistic is as follows.: (Leave any values that were given as fractions as fractions) 𝑧=𝑝̂-𝑝𝑝(1-𝑝)𝑛=( 102231800Correct -188Correct )/((188Correct ⋅(1-188Correct )) /31800Correct ) Final answer for the test statistic from technology. Round to 2 decimal places: z = 34.73IncorrectO: OBTAIN THE P-VALUE Report to 4 decimal places. It is possible when rounded that a p-value is 0.0000 P-value = .0000CorrectM: MAKE A DECISION Since the p-value Correct .04Correct , we Correct .S: STATE A CONCLUSION There Correct significant evidence to conclude Correct Correct
Solution
The question seems to be a bit jumbled, but I'll try to answer it as best as I can.
P: The parameter symbol for this problem is p, which represents the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in the state.
H: The null hypothesis (H0) would be that the proportion of children diagnosed with ASD in the state is equal to the national rate (p = 1/88). The alternative hypothesis (HA) would be that the proportion is greater than the national rate (p > 1/88).
A: The assumptions we need to check are that the sample size is large enough (n≥30), and that the expected number of successes and failures in the sample are both at least 10 (np≥10 and n(1-p)≥10).
N: The procedure we would use, assuming the conditions are met, is a one-sample z-test for a proportion.
T: The test statistic would be calculated using the formula z = (p̂ - p) / sqrt[(p(1 - p) / n], where p̂ is the sample proportion (1022/31800), p is the national proportion (1/88), and n is the sample size (31800).
O: The p-value would be obtained by looking up the test statistic in a standard normal (z) table or using a statistical software. If the p-value is less than the significance level (0.04), we would reject the null hypothesis.
M: If the p-value is less than 0.04, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the incident rate of ASD is higher in the state than nationally.
S: The conclusion would be based on the decision made in the previous step. If we rejected the null hypothesis, we would conclude that there is significant evidence to suggest that the incident rate of ASD is higher in the state than nationally. If we failed to reject the null hypothesis, we would conclude that there is not sufficient evidence to suggest a higher incident rate in the state.
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