In a survey of 2695 adults, 1446 say they have started paying bills online in th

vestirme4

vestirme4

Answered question

2021-09-18

In a survey of 2695 adults, 1446 say they have started paying bills electronically in the last year.
Construct a 99% confidence interval for the population proportion. 
Interpret your results. Choose the correct answer below. 
A) With 95% 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 
B) 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. 
C) The endpoints of the given confidence interval show that adults pay bills online 99% of the time.

Answer & Explanation

odgovoreh

odgovoreh

Skilled2021-09-19Added 107 answers

Step 1
The 100(1α)% confidence interval for the popultion proportion, p is:
(p^zα2p^(1p^)n, p^+zα2p^(1p^)n)
Here, n is the sample size, and zα/2 is the critical value of the standard normal distribution, above which, 100(α/2)%or α2  proportion of the observation lie.
Step 2
Assuming that p is the true population proportion of adults, who paid their bills online last year and the point estimate of p is p^=xn, such that, x is the number of favorable cases and n is the sample size.
As per the given query, out of 2695 adults, 1446 respondees said that they have started.
That is, n=2695 and, x=1446. That is, p^0.536549=(14462695)
If 99% of the observations must lie within an interval, the remaining 1% must lie outside the interval.
Further, due to symmetry, 0.5% will lie above the upper limit and the remaining 0.5% will lie below the lower limit of the interval.
The upper limit of the interval is thus such that, 99.5% lie below it.
As a result, zα2=z0.0052.576 [Using Excel formula: =NORM.S.INV(0.995)].
The confidence interval is,
CI=(p^±zα2p^(1p^)n)
=(0.536549±(2.576)((0.536549)(10.536549)2695))
=(0.536549±(2.576)(0.009605660255))
=(0.536549±0.02474418)
(0.512, 0.561)
Thus, the 99% confidence interval for the population proportion is 0.512<p<0.561.
Hence, it can be said with 99% confidence 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.
Correct option is B.

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