Week 9, Lab 10
Confidence intervals
Due by Mon, Oct 26
For this deliverable, you will be using the formula for calculating Confidence Intervals for sample proportions.
Remember that this formula includes three elements:
- The sample proportion.
- The Z value that corresponds to the alpha value or confidence level
- The Standard Error or Sample Deviation of the standard distribution
For the Z values, it is handy to use the following table that includes the values associated with the most common confidence levels used.
Q.1 In New York City on October 23rd, 2014, a doctor who had recently been treating Ebola patients in Guinea went to the hospital with a slight fever and was subsequently diagnosed with Ebola. Soon thereafter, an NBC 4 New York/The Wall Street Journal/Marist Poll found that 82% of New Yorkers favored a “mandatory 21-day quarantine for anyone who has come in contact with an Ebola patient”. This poll included responses of 1,042 New York adults between Oct 26th and 28th, 2014.
A) Knowing the proportion of the sample, what is the proportion in the total population of New York adults that supported a quarantine for anyone who has come into contact with an Ebola patient. Use a 95% confidence level. Write your answer.
Q.2 In 2013, the Pew Research Foundation reported that “45% 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. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval to estimate the proportion of U.S. adults who live with one or more chronic conditions. Interpret what the confidence interval tells us.
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1. c.i.= (0.796, 0.843) or (79.6%, 84.3%)
2. c.i.= (0.426, 0.473) or (42.6%, 47.3%)
The confidence interval tells us that we are 95% certain that 45% of US adults live with chronic illness.
1.Confidence Interval- 0.796 and 0.844
1A. Confidence interval-79.6% and 84.4%
2. With the confidence interval we have 95% confidence that 42.6%-47.4% of US population live with one or more chronic conditions.
1. Confidence interval = 76.6% – 84.3%, or .796 – .844
2. The confidence interval tells us that between we can be 95% certain that 42.6% and 47.3% of adults in the US live with one or more chronic conditions.
1) Confidence Interval = (0.796, 0.844) 79.6%-84.4%
2) Confidence Interval tells us that we are 95% sure that 42.6%-47.4% of US adults live with one or more chronic conditions.
1) CI= 0.796; 0.844 CI=79.6% -84.4%
2)CI= 0.45% +or -(0.012)
= 0.45+-0.02352 = 0.47352; 0.42648
The proportion is between 42.6% and 47.3%
1. C.I = 79.6%, 84.4%
2. 13.719,12.819
1. 79.6% – 84.3%
2. 42.6% – 47.3% this is the range of US adults who live with one or more chronic conditions with a 5% risk of error (the alpha level).
1). Confidence Interval= 79.6 %-84.4%. Or .796 or .844. 2). The estimate is 0.45 or 45%. According to the confidence interval, we can be 95% sure that the proportion of the adults in the United States who live with one or more chronic conditions is between 42.6%-47.34%.
1)
a) The confidence interval is ( 0.796; 0.844). Its 79.6% – 84.4%.
2) The perfect estimate is 45%. Z = 1.96 for 95% confidence level. Standard Error is 1.2%. (42.6% , 47.4%). The confidence interval tells us that we are 95% confident that US adults of 45% who live with more or one chronic circumstance is in the middle of 42.6% and 47.4%.
1.
C.I.=(.796,.844)
79.6% – 84.8%
2.
The confidence interval means we are 95% confident that of the 45% of US adults that live with chronic illness reflects the US population.