Step 1: State the Hypotheses: H0: HA: Step 2: Compute the Test statistic: (Show
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Question
Step 1: State the Hypotheses: H0: HA: Step 2: Compute the Test statistic: (Show the computations including SD) Step 3: Find the p-value: Step 4: Make the decision using the Decision Rule: (Compare the p-value with ) Step 5: Write the Conclusion (in proper English as related to the claim): ILLOWSKY Chapter 9, P. 504 73. The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. Take =0.05. (You may skip what the book is asking. Just fill in the above five steps) Chapter 9, P. 505 75. From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
Explanation / Answer
Chapter 9, P. 504 73.
Data:
n = 100
p = 0.095
p' = 0.07
Hypotheses:
Ho: p 0.095
Ha: p < 0.095
Decision Rule:
= 0.05
Critical z- score = -1.6449
Reject Ho if z < -1.6449
Test Statistic:
SE = {p (1 - p)/n} = (0.095 * (1 - 0.095)/100) = 0.0293
z = (p' - p)/SE = (0.07 - 0.095)/0.0293214938227915 = -0.8526
p- value = 0.1969
Decision (in terms of the hypotheses):
Since -0.8526 > -1.6449 we fail to reject Ho
Conclusion (in terms of the problem):
There is no sufficient evidence that the proportion is lesser than 9.5%
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