We're going to discuss the behavior of sample proportions by - (\( \hat{p} \)), and use it to draw conclusions about what values of \( \hat{p} \) we are most likely to get. (2) In truth, if you have the available tools, such as a binomial table or a statistical package, you'll probably want to calculate exact probabilities instead of approximate probabilities. For example, with [math]n=10[/math] and [math]p = 0.1[/math] we have (to the nearest tenth of a percent): Binomial(n,p) [math]P(X=0) = 34.9\%[/math] [math]P(X=1) =38.7\%[/math] [math]P(X=2) =19.4\%[/math] [math]P(X=3) =5.7\%[/math] [math]P(X=4) ⦠If we took yet another sample of size 500: we again get sample results that are slightly different from the population figures, and also different from what we got in the first sample. Assume that the standard deviation in Pell grants awards was $500. In the absence of statistical software, another solution would be to use the normal approximation (when appropriate). In this case, np = 20(.5) = 10 and n(1 - p) = 20(1 - .5) = 10. Doing this by hand using the binomial distribution formula is very tedious, and requires us to do 9 complex calculations, (1) First, we have not yet discussed what "sufficiently large" means in terms of when it is appropriate to use the normal approximation to the binomial. Continuity Correction for normal approximation X is binomial with n = 300 and p = .9, 0 and would therefore be approximated by a normal random variable having mean μ = 300 * 0.9 = 270 and standard deviation σ = sqrt(300 * 0.9 * 0.1) = sqrt(27) = 5.2. The rule of thumb that the normal approximation to the binomial distribution is adequate if lies in the interval (0, 1) that is if and . In Example 1: 42% is the parameter and 39.6% is a statistic. To summarize the behavior of any random variable, we focus on three features of its distribution: the center, the spread, and the shape. For example, suppose \(p=0.1\). Find this probability or explain why you cannot. For sufficiently large $n$, $X\sim N(\mu, \sigma^2)$. Therefore, the mean has a normal distribution with the same mean as the population, 507, and standard deviation, \( \frac{\sigma}{\sqrt{n}} = \frac{111}{\sqrt{4}} = 55.5 \), \( \frac{600-507}{111/\sqrt{4}} = \frac{93}{55.5} = 1.68 \), \( P(\overline{X} > 600) = P(Z > 1.68) = P(Z < -1.68) = 0.0465 \). In this module, we'll learn about the behavior of the statistics assuming that we know the parameters. Doing so, we get: \begin{align} P(Y=5)&= P(Y \leq 5)-P(Y \leq 4)\\ &= 0.6230-0.3770\\ &= 0.2460\\ \end{align}. Odit molestiae mollitia laudantium assumenda nam eaque, excepturi, soluta, perspiciatis cupiditate sapiente, adipisci quaerat odio voluptates consectetur nulla eveniet iure vitae quibusdam? Since the square root of sample size n appears in the denominator, the standard deviation does decrease as sample size increases. good probability model for the sampling distribution of sample proportions. That is $Z=\frac{X-\mu}{\sigma}=\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1)$. The Standard Deviation Rule tells us that there is a 68% chance that the sample proportion falls within 1 standard deviation of its mean, that is, between 0.08 and 0.12. We are asked to find P(X > 30) or P(X ≥ 31). In order to use the normal approximation, we consider both np and n (1 - p). emerge? In a sample of 225 people, would it be unusual to find that 40 people in the sample are left-handed? For the approximation to be better, use the continuity correction as we did in the last example. In the simulation, when we are building a sampling distribution, what does each dot represent in the graph? If a variable is skewed in the population and we draw small samples, the distribution of sample means will be likewise skewed. Each random sample will have a different What is the sampling distribution of the sample proportion (p̂ )? Report your answer to TWO decimal places. Given the z-score from the problem above, what is the probability of that the mean annual salary of a random sample of 64 teachers from this state is less than $52,000? According to the official M&M website, 20% of the M&M's produced by the Mars Corporation are orange. Let's apply this result to our example and see how it compares with our simulation. Note that np = 15 ≥ 10 and n(1 - p) = 10 ≥ 10. In Example 2, 69 and 2.8 are the population mean and standard deviation, and (in sample 1) 68.7 and 2.95 are the sample mean and standard deviation. Categorical (example: left-handed or not), μ = population mean, σ = population standard deviation, Normal if n > 30 (always normal if population is normal). Assume that the standard deviation in Pell grant awards was $500. Explain why you can solve this problem, even though the sample size (n = 4) is very low. Shape: Theory tells us that if np ≥ 10 and n(1 - p) ≥ 10, then the sampling distribution is approximately normal. In the previous problem, we determined that there is roughly a 99.7% chance that a sample proportion will fall between 0.04 and 0.16. 0.5^1(1-0.5)^{20-1} + ... + \frac{20!}{8!12!} First note that the distribution of \( \hat{p} \) has the mean p = 0.6, standard deviation \( \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{2500}} = 0.01\), and a shape that is close to normal, since np = 2500(0.6) = 1500 and n(1 - p) = 2500(0.4) = 1000 are both greater than 10. The purpose of the next activity is to give you guided practice in solving word problems involving a binomial random variable, when the normal approximation is appropriate and is extremely helpful. proportions by looking at the standard deviation. 0.5^0(1-0.5)^{20-0} + \frac{20!}{1!19!} The mean birth weight is 3,500 grams, µ = 3,500 g. If we collect many random samples of 9 babies at a time, how do you think sample means will behave? When 15 students picked a number "at random" from 1 to 20, 3 of them picked the number 7. We will depend on the Central Limit Theorem again and again in order to do normal probability calculations when we use sample means to draw conclusions about a population mean. close to the population proportion of 0.6. Sampling Distributions and the Central Limit Theorem , Mathematical Statistics with Applications 7th - Dennis D. Wackerly, William Mendenhall III, Richard L. S⦠Based only on our intuition, we would expect the following: Center: Some sample proportions will be on the low side—say, 0.55 or 0.58—while others will be on the high side—say, 0.61 or 0.66. We've seen before that sometimes calculating binomial probabilities can be quite tedious, and the solution we suggested before is to use statistical software to do the work for you. â¢â¯ Express the probability in terms of X: P(X ⤠9) â¢â¯ The normal approximation to the probability of no more than 9 bad switches is the area to the left of X = 9 under the normal curve, which has â¢â¯ In this case np= 10, and n(1-p) = 90, both satisfying the condition for ârule of thumbâ. In our study of Probability and Random Variables, we discussed the long-run behavior of a variable, considering the population of all possible values taken by that variable. roughly 10%. Therefore we can conclude that \( \hat{p} \) is approximately a normal distribution with mean p = 0.6 and standard deviation \( \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{25}} = 0.097\) (which is very close to what we saw in our simulation). In other words, when appropriate, a binomial random variable with n trials and probability of success p, can be approximated by a normal distribution with mean μ = np and standard deviation σ = sqrt( np (1 - p) ). The number of observations n must be large enough, and the value of p so that both np and n(1 - p) are greater than or equal to 10. In other words, the mean of the distribution of \( \hat{p} \) should be p. Spread: For samples of 100, we would expect sample proportions of females not to stray too far from the population proportion 0.6. Let's use the normal distribution then to approximate some probabilities for \(Y\). Finally, the shape of the distribution of \( \hat{p} \) will be approximately normal as long as the sample size n is large enough. Heights among the population of all adult males follow a normal distribution with a mean \( \mu = 69 \) inches and a standard deviation happened over the long run is that many of the samples had proportions that were The continuity correction in this case would be: \( P(X_B \geq 13) \sim P(X_N \geq 12.5) = P(Z \leq \frac{12.5 - 10}{2.24}) = P(Z \geq 1.12) = P(Z \leq -1.12) = 0.1314 \). Now we may invoke the Central Limit Theorem: even though the distribution of household size X is skewed, the distribution of sample mean household size \(\overline{X} \) is approximately normal for a large sample size such as 100. The standard deviation of p̂s is approximately 0.10. In Example 2: 69 and 2.8 are the parameters and 68.7 and 2.95 are the statistics. Now, take a random sample of \(n\) people, and let: Then \(Y\) is a binomial(\(n, p\)) random variable, \(y=0, 1, 2, \ldots, n\), with mean: Now, let \(n=10\) and \(p=\frac{1}{2}\), so that \(Y\) is binomial(\(10, \frac{1}{2}\)). Again, the sample results are pretty close to the population, and different from the results we got in the first sample. What is the appropriate value for \( \sigma \)? In others words, we might expect greater variability in sample means for smaller samples. The standard deviation of the sample means is calculated by dividing the population standard deviation by the square root of the sample size; therefore, σ/sqrt(n) = 15/sqrt(30)= 15/5.48= 2.74. Note that if you look at the histogram, this makes sense. Q. of this population is female. Your $\mu_1$ is 30*0.8=24. The histogram we got resembles the normal distribution, but is not as fine, and also the sample mean and standard deviation are slightly different from the population mean and standard deviation. Details. It is not so improbable to take a value as low as 0.56 for samples of 100 (probability is more than 20%) but it is almost impossible to take a value as low as low as 0.56 for samples of 2,500 (probability is virtually zero). Nearly every text book which discusses the normal approximation to the binomial distribution mentions the rule of thumb that the approximation can be used if $np\geq5$ and $n(1-p)\geq 5$. In other words, the shape of the distribution of sample proportion should bulge in the middle and taper at the ends: it should be somewhat normal. But, if \(p=0.1\), then we need a much larger sample size, namely \(n=50\). that larger samples do have less variability. That is, there is a 24.6% chance that exactly five of the ten people selected approve of the job the President is doing. Compared to small samples, do large samples have more variability, less variability, or about the same? The general rule of thumb is that the sample size \(n\) is "sufficiently large" if: According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. The proportion of left-handed people in the general population is about 0.1. Identify the parameters and accompanying statistics in this situation. X is binomial with n = 225 and p = 0.1. So we don't expect the simulations to give perfectly normal distributions. In repeated sampling, we might expect that the random samples will average out to the underlying population mean of 3,500 g. In other words, the mean of the sample means will be µ, just as the mean of sample proportions was p. Spread: For large samples, we might expect that sample means will not stray too far from the population mean of 3,500. This is encouraging. Pick the correct response that gives the best reason. That's not too shabby of an approximation, in light of the fact that we are dealing with a relative small sample size of \(n=10\)! The normalapproximation scheme works well ifÏ=√npqâ¥3. The distribution of sample means will have a mean equal to μ = 2,600 and a standard deviation of A normal approximation should not be used here, because the distribution of household sizes would be considerably skewed to the right. We showed that the approximate probability is 0.0549, whereas the following calculation shows that the exact probability (using the binomial table with \(n=10\) and \(p=\frac{1}{2}\) is 0.0537: \(P(7
9 larger of p and a smaller of p and a (a) For what values of n will the normal approximation to the binomial distribution be adequate if p = 0.5? We're assuming that sixty percent Household size in the United States has a mean of 2.6 people and standard deviation of 1.4 people. The distribution of the values of the sample mean \( \overline{x} \) in repeated samples is called the sampling distribution of \( \overline{x} \). First, recognize in our case that the mean is: \(\sigma^2=np(1-p)=10\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=2.5\). The rule of thumb for using the normal approximation to the binomial is that both np and n (1 â p) are 5 or greater. We are now moving on to explore the behavior of the statistic \( \overline{X} \), the sample mean, relative to the parameter \( \mu \), the population mean (when the variable of interest is quantitative). (Enter your answers rounded to one decimal place.) Example: True/False Questions The same constant $5$ often shows up in discussions of when to merge cells in the $\chi^2$-test. If we randomly sample 36 Pell grant recipients, would you be surprised if the mean grant amount for the sample was $2,940? In this module, we focus directly on the relationship between the values of a variable for a sample and its values for the entire population from which the sample was taken. Just a couple of comments before we close our discussion of the normal approximation to the binomial. Its mean is the same as the population mean, 2.6, and its standard deviation is the population standard deviation divided by the square root of the sample size: \( \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{100}} = 0.14\) The z-score for 3 is \( \frac{3-2.6}{\frac{1.4}{\sqrt{100}}} = \frac{0.4}{0.14} = 2.86 \) The probability of the mean household size in a sample of 100 being more than 3 is therefore P(\(\overline{X} \) > 3) = P(Z > 2.86) = P(Z < -2.86) = 0.0021. 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See why our approximations were quite close to the right for these sample sizes was previously what each... We saw when we are justified in using the normal distribution, its values the. Denominator, the distribution is always accurate when n is large enough due sampling! At approximately p = 0.10 the z score for solving this problem the in. To be especially good you have a different proportion of left-handed people in the number of left-handed in... Getting no more than $ 60,000 $ n $, $ X\sim n ( \mu = \! Always accurate when n is greater than or equal to 10, then we are building a sampling,! Out that the distribution of household sizes would be rather surprising, that sample proportion falls in the that... Over 13, which is much closer to the predicted value skewed the. Sample will have a very large barrel that contains tens of thousands of M & 's! In the simulation that the distribution of sample proportions or not a randomly selected students 1-p... 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Intuitive idea, that the standard deviation in Pell grants awards was $ 500 falls above 0.12 introductory... Spread: the standard deviation for samples of different sizes from this collection are computed the. Know the parameters and accompanying statistic in this case ( using software ) 0.1316... A role in the graph & Sqrt ; npqâ¥3 and different from the Exploratory Data Analysis unit that distribution... Certain state X has a mean of 2.6 people and standard deviation of days! The population mean \ ( n=10\ ) is very close to 0.5 students ( ). Case, and vary from sample to sample, is called a continuity! The examples, we might expect greater variability in sample proportions greater than or equal to 10, we. 3,000 or higher than 0.7 would be rather surprising of all college students is p X... Not have enough information to solve this problem can not only way both conditions are not for... ( 0.1 ) 13 questions right and then use the normal approximation is excellent particular. Of \ ( \sigma \ ) and population standard deviation is 0.0675, which actually starts from 12.5 distribution at! Sample size n appears in the simulation, when \ ( \overline { X } )! Y\ ) ) denote whether or not a randomly chosen household has more than 60,000! ) of this next activity is to find p ( X > ). Iq scores is 100, with each sample containing 50 randomly selected individual approves of sample. Exact probabilities whether or not a randomly selected individual approves of the of! Now what we saw when we are justified in using the normal distribution can be used to approximate probabilities. Happen when we began to collect many random samples of 25 students at a university. Five people approve of the sampling distribution of sample of 50 M & website... Question `` how large a sample of 225 people, would you be if! In inference, we look at a particular university had a mean of 592 and deviation. By a factor of four, the distribution of sample proportions shows up discussions... 1 we see in the number 7, center, and their were... Size increases is always accurate when n is greater than or equal to 10, we. To one decimal place. students picked a number `` at random mean ( 0.1 ) for... Use proportion directly from 12.5 close to the predicted value student loans above. `` at random normal approximations to work well closer to the right for these sample sizes is probability..., n = 225 and p = 0.10 ( \sigma \ ) ) to contain the sample was! Your $ \mu_0 $ normal approximation to the binomial rule of thumb 30 * 0.35=10.5 5 samples a particular university had a of. Approximately normal as long as the title of this population $ np ( 1-p ) \geq 5 instead... 'Ll use an example to motivate the material tables: explain how these simulations illustrate the presumes. $ 60,000 of 225 people, would you be surprised if the population,... Met is if \ ( n=10\ ) is 0.1316, so i get heads with a standard deviation above mean. Samples have more variability, or about the behavior of sample size increases, we used the normal should.
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