In 2012, Gallup asked participants if they had exercised more than 30 minutes a day for three days out of the week. Suppose that

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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?

Mathematics

1 answer:

PIT_PIT [12.4K] · Answer 1 · 2026-04-02T22:15:36+03:00

Answer:

There is a 95% likelihood that the proportion of Vermont residents who exercise for at least 30 minutes daily for three days each week falls between 55.9% and 74.7%.

Detailed explanation:

We need to determine the population proportion for Vermont.

This can be solely determined through a confidence interval since we are limited to information from a sample.

In this sample of n=100, the proportion is p=0.653.

Degrees of freedom are:

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

Calculating a 95% confidence interval, with df=99, yields a critical t value of t=1.984.

The margin of error can be computed as:

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

Thus, the limits of the confidence interval are:

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

Consequently, we conclude that there is 95% confidence that the population proportion of individuals in Vermont who exercise for at least 30 minutes a day, three times per week, is between 55.9% and 74.7%.