Question

a survey of 504 citizens found that 378 of them favor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower limit of the interval? (Round to 3 decimal digits)

Answer 1

The lower limit of the interval is 0.712 if the survey of 504 citizens found that 378 of them favor a new bill introduced by the city.

What is a confidence interval?

It is defined as the sampling distribution following an approximately normal distribution for known standard deviation.

We have:

A survey of 504 citizens found that 378 of them favor a new bill introduced by the city.

Sample proportion = p = 378/504 = 0.75

q = 1 - p = 1 - 0.75 = 0.25

$\rm SD = \sqrt{\dfrac{pq}{n}}$

$\rm SD = \sqrt{\dfrac{0.75\times0.25}{504}}$

SD = 0.01928

For 95% confidence interval Z value = 1.96

Lower limit = 0.75 - 1.96(0.01928)

= 0.712

Thus, the lower limit of the interval is 0.712 if the survey of 504 citizens found that 378 of them favor a new bill introduced by the city.

Learn more about the confidence interval here:

brainly.com/question/6654139

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