Answer:
Frank's confidence interval, (.75, .89) has to be wrong.
The explanation is presented below.
Step-by-step explanation:
The Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample proportion) ± (Margin of error)
Margin of Error is the width of the confidence interval about the mean.
So, a sample proportion of 80% indicates that Beth's interval of (.73, .87) gives a margin of error of 7%.
Making the lower limit of the interval be
(Sample proportion) - (Margin of error) = 80 - 7 = 73%
Upper limit of the interval = (Sample proportion) + (Margin of error) = 80 + 7 = 87%
Hence, this confidence interval is plausible for the survey carried out.
But Frank's interval of (.75, .89) doesn't work with the definition of a confidence interval given above. The range has to be equally spread around the mean with (lower limit, upper limit). The lower limit cannot be 75 and the upper limit 89.
If the lower limit is truly 75, the margin of error has to be (80 - 75), 5%. And the corresponding upper limit should be (80 + 5), 85%.
Frank's interval as (.75, .85) would have been more plausible.
And if the upper limit is truly 89%, the fiitting lower limit, judging by the distance of the upper limit from the sample proportion (89 - 80 = 9%), should be (80 - 9), 71%. Frank's interval as (.71, .89) would also have been more plausible.
Hope this Helps!!!
) you may want to used the distance formula if you are not given any other points (with the exception of the vertex, asymptote or roots... etc).