3. According to the Centers for Disease Control and Preventions, 18.6% of the U.

Question

3. According to the Centers for Disease Control and Preventions, 18.6% of the U.S. population smoked in 2008. In 2016, a random sample of 1000 Americans was selected, 114 of whom smoked a. Construct a 90% confidence interval to estimate the actual proportion of people who smoked in the United States in 2016. Construct a 96% confidence interval to estimate the actual proportion of people who smoked in the United States in 2016. Is there any evidence that this proportion has changed since 2008 based on this sample? b. c.

Explanation / Answer

3) Sample size = n = 1000

x = 114

a) In this part, we have to construct a 90% confidence interval to estimate the actual proportion of people who smoked in the United States in 2016.

By using TI-84 we can solve easily

Path is Click on STAT -----> TESTS --------> 1-PropZInt -----------> Enter values ---->

x: 114

n: 1000

C-Level: 0.90

Calculate

We get confidence interval ( 0.09747 , 0.13053)

b) In this part, we have to construct a 96% confidence interval to estimate the actual proportion of people who smoked in the United States in 2016.

By using TI-84 we can solve easily

Path is Click on STAT -----> TESTS --------> 1-PropZInt -----------> Enter values ---->

x: 114

n: 1000

C-Level: 0.9

Calculate

We get confidence interval ( 0.09336 , 0.13464)

c)

There is no evidence that the proportion has changed since 2008 based on given sample because 0.186 does not belong to both intervals.

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