In a survey of 3132 adults,1424 say they have started paying bills online in the

Question

In a survey of 3132 adults,1424 say they have started paying bills online in the last year.

Construct a 99% confidence interval for the population proportion. Interpret the results.

A 99% confidence interval for the population proportion is

(Round to three decimal places as needed.)

Interpret your results. Choose the correct answer below.

A. With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

B. The endpoints of the given confidence interval show that adults pay bills online 99% of the time.

C. With 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Explanation / Answer

Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=1424
Sample Size(n)=3132
Sample proportion = x/n =0.455
Confidence Interval = [ 0.455 ±Z a/2 ( Sqrt ( 0.455*0.545) /3132)]
= [ 0.455 - 2.576* Sqrt(0) , 0.455 + 2.58* Sqrt(0) ]
= [ 0.432,0.478]
Interpretations:
1) We are 99% sure that the interval [0.432 , 0.478 ] contains the true population proportion
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion  
C. With 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

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