Hope one of you can help:
I've got a general quesiton on finding confidence intervals for the porportion of a population. I dont understand how it is possible to calculate it only taking into account information about the sample. We know nothing about the relationship between the sample and the population. When we do a simillar calculation for the distribution of the mean we know that the there is a relationship between the sd of the population and the sd of the means and that allows us to take information from the sample and use it to inform us about the behaviour of the population.
But that is not true for poportions. The sd of a sample is not related to the sd of the population as we can see by looking at its formula.
Am I making any sense?
thanks alot for your time
regards
Pedro
The drug Viagra became available in the U.S. in May, 1998, in the wake of an advertising campaign that was unprecedented in scope and intensity. A Gallup poll found that by the end of the first week in May, 643 out of a random sample of 1,005 adults were aware that Viagra was an impotency medication (based on "Viagra A Popular Hit," a Gallup poll analysis by Lydia Saad, May 1998).
Let's estimate the proportion p of all adults in the U.S. who by the end of the first week of May 1998 were already aware of Viagra and its purpose by setting up a 95% confidence interval for p.
We first need to calculate the sample proportion [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT]. Out of 1,005 sampled adults, 643 knew what Viagra is used for, so [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]643[/FONT][FONT=MathJax_Main]1005[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64
[/FONT]
Therefore,
A 95% confidence interval for p is [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Size1]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Size1])[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Size1]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Size1])[/FONT][FONT=MathJax_Main]1005[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]03[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]61[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]67[/FONT][FONT=MathJax_Main])[/FONT]
We can be 95% sure that the proportion of all U.S. adults who were already familiar with Viagra by that time was between .61 and .67 (or 61% and 67%).
The fact that the margin of error equals .03 says we can be 95% confident that unknown population proportion p is within .03 (3%) of the observed sample proportion .64 (64%). In other words, we are 95% confident that 64% is "off" by no more than 3%.
I've got a general quesiton on finding confidence intervals for the porportion of a population. I dont understand how it is possible to calculate it only taking into account information about the sample. We know nothing about the relationship between the sample and the population. When we do a simillar calculation for the distribution of the mean we know that the there is a relationship between the sd of the population and the sd of the means and that allows us to take information from the sample and use it to inform us about the behaviour of the population.
But that is not true for poportions. The sd of a sample is not related to the sd of the population as we can see by looking at its formula.
Am I making any sense?
thanks alot for your time
regards
Pedro
The drug Viagra became available in the U.S. in May, 1998, in the wake of an advertising campaign that was unprecedented in scope and intensity. A Gallup poll found that by the end of the first week in May, 643 out of a random sample of 1,005 adults were aware that Viagra was an impotency medication (based on "Viagra A Popular Hit," a Gallup poll analysis by Lydia Saad, May 1998).
Let's estimate the proportion p of all adults in the U.S. who by the end of the first week of May 1998 were already aware of Viagra and its purpose by setting up a 95% confidence interval for p.
We first need to calculate the sample proportion [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT]. Out of 1,005 sampled adults, 643 knew what Viagra is used for, so [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]643[/FONT][FONT=MathJax_Main]1005[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64
[/FONT]
Therefore,
A 95% confidence interval for p is [FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Size1]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]p[/FONT][FONT=MathJax_Size1]ˆ[/FONT][FONT=MathJax_Size1])[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Size1]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Size1])[/FONT][FONT=MathJax_Main]1005[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size2]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]64[/FONT][FONT=MathJax_Main]±[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]03[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]61[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main].[/FONT][FONT=MathJax_Main]67[/FONT][FONT=MathJax_Main])[/FONT]
We can be 95% sure that the proportion of all U.S. adults who were already familiar with Viagra by that time was between .61 and .67 (or 61% and 67%).
The fact that the margin of error equals .03 says we can be 95% confident that unknown population proportion p is within .03 (3%) of the observed sample proportion .64 (64%). In other words, we are 95% confident that 64% is "off" by no more than 3%.