Question

In 2012, Gallup asked participants if they had exercised more than 30 minutes a day for three days out of the week. Suppose that random samples of 100 respondents were selected from both Vermont and Hawaii. From the survey, Vermont had 65.3% who said yes and Hawaii had 62.2% who said yes. What is the value of the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week?

Answer 1

Answer:

There is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.

Step-by-step explanation:

We have to answer the population proportion for Vermont.

We can only do it by a confidence interval, as we only have information from a sample.

This sample, of size n=100, has a proportion p=0.653.

The degrees of freedom are:

$df=n-1=100-1=99$

We will calculate a 95% confidence interval, which for df=99 has a critical value of t of t=1.984.

The margin of error can be calculated as:

$E=t*\sigma_p=t\sqrt{\dfrac{p(1-p)}{n}}=1.984\sqrt{\dfrac{0.653*0.347}{100}}\\\\\\E=1.984*\sqrt{0.00226}=1.984*0.0476=0.094$

Then, the upper and lower bounds of the confidence interval are:

$LL=p-E=0.653-0.094=0.559\\\\UL=p+E=0.653+0.094=0.747$

Then, we can say that there is 95% confidence that the population proportion of people from Vermont who exercised for at least 30 minutes a day 3 days a week is between 55.9% and 74.7%.