g In 2013, the Pew Research Foundation reported that 43% of U. S. adults report that they live with one or more chronic conditions. However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.7%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U. S. adults who live with one or more chronic conditions. Report your answer as a percentage rounded to 3 decimal places. We are 95% confident that the true proportion of U. S. adults who live with one or more chronic conditions is between % and % Submit All Parts Question 6

We are 95% confident that the true proportion of U.S. adults who live with one or more chronic conditions is between 39.7% and 46.33%

Step-by-step explanation:

From the question we are told that

  The  sample  proportion is

   The  standard error is

Given that the confidence level is 95%  then the level of significance is mathematically represented as  

=>  

Generally from the normal distribution table the critical value  of   is  

Generally the margin of error is mathematically represented as

=>    

=>    

Generally 95% confidence interval is mathematically represented as

=>  

=>    

Converting to percentage

y=1/4x-4

step-by-step explanation:

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional