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Anuta_ua [19.1K]
3 years ago
7

In 2013, the Pew Research Foundation reported that "45% of U.S. adultsreport that they live with one or more chronic conditions"

, and the standard error for this estimate is 1.2%.Identify each of the following statements as true or false. Provide an explanation to justify each of youranswers.(a) We can say with certainty that the confidence interval from Exercise 5.7 contains the true percentageof U.S. adults who suffer from a chronic illness.(b) If we repeated this study 1,000 times and constructed a 95% confidence interval for each study, thenapproximately 950 of those confidence intervals would contain the true fraction of U.S. adults who sufferfrom chronic illnesses.(c) The poll provides statistically significant evidence (at theα= 0.05 level) that the percentage of U.S.adults who suffer from chronic illnesses is below 50%.(d) Since the standard error is 1.2%, only 1.2% of people in the study communicated uncertainty abouttheir answer.
Mathematics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

b) True , this is right. But this is not for 1000 alone for any sample size of large number randomly drawn we can expect 95%

Step-by-step explanation:

Given that in 2013, the Pew Research Foundation reported that "45% of U.S. adults report that they live with one or more chronic conditions", and the standard error for this estimate is 1.2%

Let us check one by one answers given

a) False.  Because we can say 95% or 90% confident only that the true percent of US adults cannot be true.

b) True , this is right. But this is not for 1000 alone for any sample size of large number randomly drawn we can expect 95%

c) False this does not give whether less than 50% or not unless we know the confidence interval range.

d) False, std error is a measure of variation of all the items from the mean.

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