Calculating Listeners Tuning in 8-9am: Least/Greatest #s

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Discussion Overview

The discussion revolves around calculating the least and greatest number of listeners who tune into a radio program between 8 and 9 am, based on a claim of 55% listenership and a survey of 1000 listeners. The focus includes the application of margin of error formulas and confidence intervals.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a calculation using a margin of error formula but expresses uncertainty about arriving at the correct answer.
  • Another participant suggests that the question involves finding a confidence interval and asks for clarification on how the margin of error formula is applied in the context of the course material.
  • A different participant confirms that the answer of 519 to 581 is provided in the textbook and discusses the margin of error formula, noting the lack of population standard deviation.
  • Another participant reiterates the textbook answer and affirms that the original formula used by the first participant was correct, despite initial confusion.

Areas of Agreement / Disagreement

Participants generally agree on the answer of 519 to 581 as stated in the textbook, but there is some uncertainty regarding the application of the margin of error formula and the interpretation of the results.

Contextual Notes

There is mention of different approaches to calculating margin of error and confidence intervals, as well as the absence of population standard deviation, which may affect the application of certain formulas.

markosheehan
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a radio station claims 55% of all listeners tune into their program between 8 and 9 am. To check this claim a competing radio station does a survey of a random sample of listeners in the 8 to 9 am time slot. the survey indicates the claim can not be rejected. if the sample size is 1000 people determine the least and greatest number of listeners that could of said they listen to the radio station.

this is what i have tried but i am not getting the right answer.

p=.55

using the margin of error formula. 1/ √1000 = .03126
so I would of thought the answer would be 1.55 +/- .03126 multiplied by a thousand. this does not give the right answer. the right answer is 519 to 581
 
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markosheehan said:
a radio station claims 55% of all listeners tune into their program between 8 and 9 am. To check this claim a competing radio station does a survey of a random sample of listeners in the 8 to 9 am time slot. the survey indicates the claim can not be rejected. if the sample size is 1000 people determine the least and greatest number of listeners that could of said they listen to the radio station.

this is what i have tried but i am not getting the right answer.

p=.55

using the margin of error formula. 1/ √1000 = .03126
so I would of thought the answer would be 1.55 +/- .03126 multiplied by a thousand. this does not give the right answer. the right answer is 519 to 581

Hi markosheehan,

So this question basically comes down to finding a reasonable interval that we would expect the numbers to fluctuate between. This is often called a "confidence interval" and corresponds to a "fail to reject region" of the null hypothesis. I'm not sure how the material is being presented in your course but can you tell me how you've been told to apply the margin of error formula?

Basically no matter what this is going to look like this: $555 \pm \text{noise}$. Different teachers and courses approach this differently so I'm curious as to what you've been told to do. The answer you provided is not symmetric. Can you confirm that the answer should be 519 to 581? If it were 529 to 581 then that would be symmetric, that is $555 \pm 26$.
 
519, 581 is the answer at the back of the book.

Usually we use the margin of error formula
E=1/✓n

We have also used the margin of error (for c% confidence interval) E=z* standard deviation of proportion where z is 1.96 at the 95 % confidence. I'm not too sure how this formula works or what it is even used for. But anyway it can not be used here due to the reason we are not given the standard deviation of the population.
 
markosheehan said:
519, 581 is the answer at the back of the book.

Usually we use the margin of error formula
E=1/✓n

We have also used the margin of error (for c% confidence interval) E=z* standard deviation of proportion where z is 1.96 at the 95 % confidence. I'm not too sure how this formula works or what it is even used for. But anyway it can not be used here due to the reason we are not given the standard deviation of the population.

We don't know the standard deviation of the population, but we do know that SD of the sample. See this page. It looks like this isn't the way you are going though for this homework.

Your original formula was:
[math]\left(55 \pm \frac{1}{\sqrt{1000}} \right) \times 1000 = (519,581)[/math]

So actually you had it right the first time. :)
 
thanks
 

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