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Sample Proportion

Sample Proportion.

neil
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Sample Proportion

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  1. Sample Proportion

  2. Say there is a huge company with a boat load of employees. If we could talk to all the employees we could ask each one whether or not they participated in a management training program with the company. We would then be able to calculate the population proportion (or percentage) that said yes. But, remember we usually do not do a census (talk to everyone), we do a sample and try to learn about the population based on the sample. From the sample we could calculate a sample proportion, called p hat. In a theoretical sense, p hat will vary from sample to sample and thus has a distribution.

  3. It turns out that experts have figured out the properties of the distribution of p hat. It is similar to the central limit theorem we saw with the sampling distribution of sample means. The sampling distribution of p hat 1. Is a normal distribution 2. Has expected value (mean) equal to the population proportion p, 3. Has standard deviation, what is called the standard error of the proportion, equal to the square root of some formula. That formula is [p(1 – p)]/n, where p is the population proportion.

  4. Let’s say we are told that the population proportion is .4 in a certain situation. (You would think we could quit if we know about the population – we really could but we are trying to learn some patterns of thinking.) Say a sample of 200 is taken. The sample proportion could be calculated – that is not what this problem is about. What is the probability that the sample proportion will be within + or - .03 of the population proportion? In other words, what is the probability of being in the neighborhood from .37 to .43? To answer this we have to use the three properties I mentioned before. We will use the normal distribution, with mean =.4 and standard error = square root of {[.4(.6)]/200} = square root(.0012) = .0346

  5. At this stage we can go use Excel with the NORMSDIST function which is based on z values, or we could use NORMDIST, which is based on the value of interest, the mean,and the standard error. .37 has a z = (.37 - .4)/.0346 = -.87. .43 has a z = .87 This is the area we want and this is my sketch of the answer. From the table we have .8078 - .1922 = .6156 Sampling dist. of p .37 .4 .43

  6. Problem where p = .4 and n = 50 a. p hat= 15/50 = .30 b. Standard error = sqrt[.4(.6)/50] = .0693 In part a I am saying that out of 50 people in a sample 15 had a characteristic that made the sample proportion .3

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