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# Standard Error For The Sampling Distribution Of Proportion

repeatedly randomly drawn from a population, and the proportion of successes in each sample is

## Standard Deviation Of Sample Proportion

recorded ($$\widehat{p}$$),the distribution of the sample proportions (i.e., the sampling sample proportion formula distirbution) can be approximated by a normal distribution given that both $$n \times p sample proportion definition \geq 10$$ and $$n \times (1-p) \geq 10$$. This is known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15

## Sample Proportion Calculator

instead of 10.The mean of the distribution of sample proportions is equal to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. The symbol $$\sigma _{\widehat p}$$ is

## P Hat Calculator

also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3

taken, the distribution of $$\hat{p}$$ is said to approximate a normal curve distribution. Alternatively, this can be assumed if BOTH n × p and n × (1 p hat symbol - p) are at least 10. [SPECIAL NOTE: Some textbooks use 15

## P Hat Formula

instead of 10 believing that 10 is to liberal. We will use 10 for our discussions.] Using this, probability of sample proportion calculator we can estimate the true population proportion, p, by $$\hat{p}$$ and the true standard deviation of p by $$s.e.(\hat{p})=\sqrt{\frac {p(1-p)}{n}}$$, where s.e.( $$\hat{p}$$) is interpreted as the standard error https://onlinecourses.science.psu.edu/stat200/node/43 of $$\hat{p}$$. Probabilities about the number X of successes in a binomial situation are the same as probabilities about corresponding proportions. In general, if np ≥ 10 and n(1- p) ≥ 10, the sampling distribution of$$\hat{p}$$ is about normal with mean of p and standard error $$s.e.(\hat{p})=\sqrt{\frac {p(1-p)}{n}}$$. Example. Suppose the proportion of all college students who have used marijuana https://onlinecourses.science.psu.edu/stat800/node/35 in the past 6 months is p = .40. For a class of size N = 200, representative of all college students on use of marijuana, what is the chance that the proportion of students who have used mj in the past 6 months is less than .32 (or 32%)? NOTE: This would imply that 32% of the sample students said "yes" to having used marijuana, or 64 of the 200 said "yes". This means the sample proportion $$\hat{p}$$ is 64/200 or 32%Solution. The mean of the sample proportion$$\hat{p}$$ is p and the standard error of$$\hat{p}$$ is $$s.e.(\hat{p})=\sqrt{\frac {p(1-p)}{n}}$$. For this marijuana example, we are given that p = .4. We then determine $$s.e.(\hat{p})=\sqrt{\frac {p(1-p)}{n}}=\sqrt{\frac{0.4(1-0.4)}{200}}=0.0346$$. So, the sample proportion$$\hat{p}$$ is about normal with mean p = .40 and SE($$\hat{p}$$) = 0.0346.The z-score for .32 is z = (.32 - .40) / 0.0346 = -2.31. Then using Standard Normal Table Prob($$\hat{p}$$ < .32) = Prob(Z <. -2.31) = 0.0104. Question to ponder: If you observed a sample proportion of .32 would you believe a claim that 40

distribution of a sample proportion compute probabilities of a sample proportion For https://faculty.elgin.edu/dkernler/statistics/ch08/8-2.html a quick overview of this section, feel free to watch this short video summary: The Sample Proportion Consider these recent headlines: Hispanics See Their Situation in U.S. Deteriorating Half http://www.milefoot.com/math/stat/samp-proportions.htm (50%) of all Latinos say that the situation of Latinos in this country is worse now than it was a year ago, according to a new nationwide survey of sample proportion 2,015 Hispanic adults conducted by the Pew Hispanic Center. (Source: Pew Research) Automatic enrollment in 401(k) doesn't take care of everything Never got around to signing up for the company retirement plan? The boss may have done it for you. Forty-two percent of employers with 401(k) plans automatically enroll new or existing employees in the plans, nearly double the of sample proportion 23 percent from 2006, according to estimates from the 2008 401(k) Benchmarking Survey by the International Foundation of Employee Benefit Plans and Deloitte Consulting. The survey polled 436 employers with workforces of all sizes. (Source: Chicago Tribune) Stem cell, marijuana proposals lead in Mich. poll A recent poll shows voter support leading opposition for ballot proposals to loosen Michigan's restrictions on embryonic stem cell research and allow medical use of marijuana. The EPIC-MRA poll conducted for The Detroit News and television stations WXYZ, WILX, WOOD and WJRT found 50 percent of likely Michigan voters support the stem cell proposal, 32 percent against and 18 percent undecided. (Source: Associated Press) These three articles all have something in common - they're referring to sample proportions - 50% of all Latinos, 42% of employers, and 50% of likely Michigan voters, respectively, in the three articles above. Proportions are the number with that certain characteristics divided by the sample size. In general, if we let x = the number with the specific characteristic, then the sample proporti

often think of a mathematical proportion as an equality of two ratios, in statistics the proportion is a percentage of a total in which a certain characteristic is observed. If a population has size $N$, and the characteristic occurs $x$ times in that population, then the population proportion is given by $p = \dfrac{x}{N}$. If a sample of size $n$ is obtained, and the characteristic occurs $x$ times in the sample, then the proportion in that sample is given by $\hat{p} = \dfrac{x}{n}$. There is a connection between these formulas for the proportion and a binomial distribution. In fact, the formula $p = \dfrac{x}{N}$, if solved for $x$, gives the expected value of the number of successes for a binomial distribution, $x = Np$. Looking further, we see that if the observed characteristic is considered as a success, then not observing it is a failure. The probability of a success is $p$. If individuals are randomly selected from a very large population, then we can assume that the selections are independent, and that the probabilities will be constant. Therefore, all of the conditions of the binomial distribution are met for the variable $x$. So what is the expected value of a sample proportion, $E(\hat{p})$? The binomial result leads us to the answer. $E(\hat{p}) = E \left( \dfrac{x}{n} \right) = \dfrac1n E(X) = \dfrac1n (np) = p$ Similarly, we can find the variance in a population of sample proportions. $Var(\hat{p}) = Var \left( \dfrac{x}{n} \right) = \dfrac{1}{n^2} Var(X) = \dfrac{1}{n^2} np(1-p) = \dfrac{p(1-p)}{n}$ And from this result, we can easily obtain the standard deviation. Therefore, we have the following parameters for a distribution of sample proportions. $\mu_{\hat{p}} = p$ $\sigma_{\hat{p}} = \sqrt{ \dfrac{p(1-p)}{n} }$ If the values $np$ and $n(1-p)$ are both at least 5, then the binomial distribution of $X$ will be approximately normal, and it will follow that the sampling distribution of the proportions will also be approximately normal, and can be standardized with the formula $z = \dfrac{\hat{p} - p}{\sigma_{\hat{p}}}$. An Example Suppose the true value of the president's approval rating is 56%. Find the probability that a sample of 1200 people would find a proportion between 53% and 58%. The standard deviation of the sample proportions is $\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} = \sqrt{\dfrac{(0.56)(0.44)}{1200}} \approx 0.0143$ The z-scores are $z = \dfrac{0.53-0.56}{0.0143} \approx -2.10$ and $z = \dfrac{0.58-0.56}{0.0453} \approx 1.40$. Computing the probability using the standard normal distribution,

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