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

AI Thread Summary
A radio station claims that 55% of listeners tune in between 8 and 9 am, prompting a competing station to survey 1,000 listeners to verify this claim. The survey results indicate that the claim cannot be rejected, and the correct range of listeners who could have said they listen is between 519 and 581. The margin of error was calculated using the formula E = 1/√n, leading to the conclusion that the initial calculations were on the right track. The discussion highlights the importance of understanding confidence intervals and the application of margin of error in statistical analysis. Ultimately, the correct interpretation confirms the range of listeners aligns with the station's claim.
markosheehan
Messages
133
Reaction score
0
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
 
Mathematics news on Phys.org
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
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
5
Views
9K
Back
Top