Central limit theorem and estimates of probability

In summary: The central limit theorem states that the sample proportion will be approximately normally distributed with a large enough number of observations. Therefore, by using the z-value and the normal distribution, it is possible to calculate the probability that the sample proportion is greater than .5. This probability is 0.95.
  • #1
lotsofmoxie
16
0

Homework Statement


Assume five hundred people are given one question to answer - the question can be answered with a yes or no. Let p =the fraction of the population that answers yes. Give an estimate for the probability that the percent of yes answers in the five hundred person sample is bigger than that of in the whole population (p) by more than 5 %

Homework Equations



central limit theorem

p is NOT provided.

The Attempt at a Solution



N=.55*500
n=500

using a typical normal approximation to the binomial we get

P(X>N)=1-CDFOFNORMALDIST[(N-np)/sqrt(n*p*(1-p))]

How can we go farther than this if we don't have an explicit value for p?[/B]
 
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  • #2
You have to use your understanding of how the distribution of the sample compares with the distribution in an entire population.
Be careful, you have used the variable "p" to mean at least two different things.
 
  • #3
What can we say about the population distribution besides it is binomial and about the sample distribution besides its average is approximately normal with large enough n?
 
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  • #4
I'm really not seeing how it is possible to do this problem without knowing the probability that a particular person answers yes or no. You need this to calculate an exact Z value, and how large or small it is makes a huge difference to the answer. My gut tells me that we are just supposed to assume that this is equal to .5 (equal chance of yes or no), but maybe I'm just stupid and missing something big conceptually.
 
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  • #5
I'm really not seeing how it is possible to do this problem without knowing the probability that a particular person answers yes or no. You need this to calculate an exact Z value, and how large or small it is makes a huge difference to the answer - from what I can see, the central limit theorem explicitly requires it in order to use the normal distribution at all. My gut tells me that we are just supposed to assume that this is equal to .5 (equal chance of yes or no), but maybe I'm just stupid and missing something big conceptually.
 
  • #6
lotsofmoxie said:
I'm really not seeing how it is possible to do this problem without knowing the probability that a particular person answers yes or no. You need this to calculate an exact Z value, and how large or small it is makes a huge difference to the answer - from what I can see, the central limit theorem explicitly requires it in order to use the normal distribution at all. My gut tells me that we are just supposed to assume that this is equal to .5 (equal chance of yes or no), but maybe I'm just stupid and missing something big conceptually.
As Simon said you are using p for two different things. You can use the fact that p(1-p) (where 0<p<1) has a maximum value(find it). You can use it to get a bound for your answer.
 
  • #7
Hmmm. So the max value of p(1-p) (if I'm using p = probability that a given person answers yes to the question) would be 1/2. So I can plug that into my z value?
 
  • #8
lotsofmoxie said:
Hmmm. So the max value of p(1-p) (if I'm using p = probability that a given person answers yes to the question) would be 1/2. So I can plug that into my z value?
It's not 1/2. Check your calculation.
 
  • #9
Err sorry, typo. It's 1/4.

so we get P(X>275) >= 1-cdfnorm[(275-275*p)/(sqrt(n/4))

what about the p in the numerator? Are we trying to get a lower or upper bound?
 
  • #10
Is the answer zero? I can post my reasoning if its forum policy or something
 
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  • #11
Try a hypothesis test for population proportions, which relates the actual population proportion to the sample proportion. You first find the standard deviation ## \sigma ## of the sampling distribution of the population proportion P :

## \sigma = \sqrt {\frac {P(1-P)}{n}} ## and then use the normally-distributed statistic z:

## z =\frac {p-P}{\sigma} ## , where p is the sample proportion. This gives you the probability of obtaining a sample proportion value p in a population with actual proportion P.

From this calculate the percentile value associated with the ##z##-value you got. You want to find the probability that p-P>0.05.
 

Related to Central limit theorem and estimates of probability

1. What is the central limit theorem?

The central limit theorem is a fundamental concept in statistics that states that the sampling distribution of the mean of any independent, random variable will approximate a normal distribution as the sample size increases.

2. How is the central limit theorem used in practice?

The central limit theorem is used in practice to estimate population parameters, such as the mean or variance, from a sample of data. It allows us to make inferences about a population based on a sample, and is used in a wide range of fields including economics, psychology, and biology.

3. What are the assumptions of the central limit theorem?

The central limit theorem assumes that the individual observations in the sample are independent and identically distributed, and that the sample size is sufficiently large.

4. How does the central limit theorem relate to the concept of standard error?

The central limit theorem is closely related to the concept of standard error, which is the measure of the variability of the sample mean from one sample to another. The central limit theorem states that as the sample size increases, the standard error of the sample mean will decrease, indicating a more precise estimate of the population mean.

5. Can the central limit theorem be applied to any type of data?

The central limit theorem can be applied to any type of data, as long as the assumptions are met. However, it is most accurate with continuous data and can be less accurate with discrete or skewed data. In these cases, alternative methods may be used to estimate population parameters.

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