assume that in a large population, the probability

A random selection of 10 playing cards from a deck of 52 cards implies that. Remember that the area to the right of $z_{0.025}$ is $0.025$ and the area to the left of $z_{0.025}$ is $0.975$. This page titled 6.5: Estimating Population Proportion is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax. X Introductory Statistics (Shafer and Zhang), { "7.01:_Large_Sample_Estimation_of_a_Population_Mean" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.02:_Small_Sample_Estimation_of_a_Population_Mean" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.03:_Large_Sample_Estimation_of_a_Population_Proportion" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.04:_Sample_Size_Considerations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.E:_Estimation_(Exercises)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Introduction_to_Statistics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Descriptive_Statistics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Basic_Concepts_of_Probability" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Discrete_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Continuous_Random_Variables" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Sampling_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Estimation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Testing_Hypotheses" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Two-Sample_Problems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Correlation_and_Regression" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "11:_Chi-Square_Tests_and_F-Tests" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 7.2: Small Sample Estimation of a Population Mean, [ "article:topic", "Student\u2019s t-distribution", "showtoc:no", "license:ccbyncsa", "program:hidden", "licenseversion:30", "source@https://2012books.lardbucket.org/books/beginning-statistics", "authorname:anonymous" ], https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FBookshelves%2FIntroductory_Statistics%2FIntroductory_Statistics_(Shafer_and_Zhang)%2F07%253A_Estimation%2F7.02%253A_Small_Sample_Estimation_of_a_Population_Mean, $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}$ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$, Small Sample $ 100(1)\%$ Confidence Interval for a Population Mean, 7.1: Large Sample Estimation of a Population Mean, 7.3: Large Sample Estimation of a Population Proportion, source@https://2012books.lardbucket.org/books/beginning-statistics. We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%. lowervalueofthearea,uppervalueofthearea,mean, \[\mu _{\hat{p}}=p=0.38\; \text{and}\; \sigma _{\hat{P}}=\sqrt{\frac{pq}{n}}=\sqrt{\frac{(0.38)(0.62)}{900}}=0.01618 \nonumber \], Then $3\sigma _{\hat{P}}=3(0.01618)=0.04854\approx 0.05$ so, \[\left [ \hat{p} - 3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\, \hat{p}+3\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right ]=[0.38-0.05,0.38+0.05]=[0.33,0.43] \nonumber \]. Calculate the sample size required to estimate a population proportion given a desired confidence level and margin of error. If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones. The Central Limit Theorem has an analogue for the population proportion $\hat{p}$. There are many online calculators that can be used to compute the confidence intervals. \[E=z_{\alpha /2}\left ( \frac{\sigma }{\sqrt{n}} \right ) \nonumber \], \[E=z_{\alpha /2}\left ( \frac{s}{\sqrt{n}} \right ) \nonumber \]. Jun 23, 2022 OpenStax. For confidence level $90\%$, $\alpha =1-0.90=0.10$, so $z_{\alpha /2}=z_{0.05}$. samplesize are not subject to the Creative Commons license and may not be reproduced without the prior and express written (a) the random sample of 10 cards accurately represents the important features of the whole deck. var edata = JSON.parse(e.originalEvent.data); Use $p=0.90$, corresponding to the assumption that the retailers claim is valid. To understand how to apply additional formulas for a confidence interval for a population mean. 2, 25 The sample proportion is the number $x$ of orders that are shipped within $12$ hours divided by the number $n$ of orders in the sample: Since $p=0.90$, $q=1-p=0.10$, and $n=121$. ). 25 Suppose a . There are formulas for the mean $_{\hat{P}}$, and standard deviation $_{\hat{P}}$ of the sample proportion. Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters. As sample sizes increase, the distribution of means more closely follows the normal distribution. Since the sample size is $n = 15$, there are $n1=14$ degrees of freedom. This gives us the following confidence interval formulas. Often sampling is done in order to estimate the proportion of a population that has a specific characteristic, such as the proportion of all items coming off an assembly line that are defective or the proportion of all people entering a retail store who make a purchase before leaving. For example, when CL = 0.95, = 0.05 and 2 = 0.025; we write z2=z0.025. Take a sample of size n = 100. The sample proportion is a random variable $\hat{P}$. Construct a $95\%$ confidence interval for the population mean, and interpret its meaning. b. Repeat the previous question, this time using the Poisson distribution to find an approximate probability. Assuming the retailer's claim is true, find the probability that a sample of size $121$ would produce a sample proportion so low as was observed in this . Figure $\PageIndex{1}$ shows that when $p = 0.1$, a sample of size $15$ is too small but a sample of size $100$ is acceptable. The confidence interval formulas in the previous section are based on the Central Limit Theorem, the statement that for large samples $\overline{X}$ is normally distributed with mean $\mu$ and standard deviation $\sigma /\sqrt{n}$. Accessibility StatementFor more information contact us atinfo@libretexts.org. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. De nition 2 (Convergence in probability). ( 9 votes) Upvote Flag Rishabh Chopra 4 years ago How does the 10% rule make sense? The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. Thus, \[\begin{align} \overline{x} &= t_{/2} \left( \dfrac{s}{\sqrt{n}}\right) \\ &=2.71 1.796 \left( \dfrac{0.51}{\sqrt{12}} \right) \\ &=2.71 0.26 \end{align} \nonumber \], One may be $90\%$ confident that the true average GPA of all students at the university is contained in the interval, \[(2.710.26,2.71+0.26)=(2.45,2.97). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. $(window).on("message", function(e) { A sample is large if the interval $\left [ p-3\sigma _{\hat{p}},\, p+3\sigma _{\hat{p}} \right ]$ lies wholly within the interval $[0,1]$. ) }); ). The population proportion is denoted $p$ and the sample proportion is denoted $\hat{p}$. The central limit theorem for sample means says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate their means, those means tend to follow a normal distribution (the sampling distribution). Find the probability that the sample mean is between eight minutes and 8.5 minutes. . Chapter 8, Population, Samples, and Probability Video - Numerade Population & Demographics. To form a proportion, take $X$, the random variable for the number of successes and divide it by $n$, the number of trials (or the sample size). 25 Of the 250 surveyed, 98 reported owning a tablet. Arrow down to C-Level and enter 0.90. If the cans are filled so that = 16.00 ounces (as labeled) and = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. The amounts in a sample are measured and the statistics are n = 34, samplesize A sample size of n = 60 is drawn randomly from the population. Check the Real Estate section in your local. Clearly the proportion of the population with the special characteristic is the proportion of the numerical population that are ones; in symbols, \[p=\dfrac{\text{number of 1s}}{N} \nonumber \], But of course the sum of all the zeros and ones is simply the number of ones, so the mean  of the numerical population is, \[=\dfrac{ \sum x}{N}= \dfrac{\text{number of 1s}}{N} \nonumber \]. What are the properties of normal distributions? 25 To put it more formally, if you draw random samples of size n, the distribution of the random variable How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones? it is appropriate to use the normal distribution to compute probabilities related to the sample proportion $\hat{P}$. Support your answer. The probability that sample proportion < 0.7 is the tops of all the rectangles below 0.7 summed up for the sampling distribution. If 5 people are selected at random from the population, what is the probability that at least 1 of the people selected will always take medicine as prescribed? A sample is considered large when $n\geq 30$. Chapter 6-4 Flashcards | Quizlet is the value of . = 0.9977. As also indicated by the figure, as the sample size $n$ increases, Students $t$-distribution ever more closely resembles the standard normal distribution. See Answer Question: Assume that in a large city, the distribution of the number of children per a family is as follows. ) Since $CL = 0.96$, we know $\alpha = 1 0.96 = 0.04$ and $\dfrac{\alpha}{2} = 0.02$. Use $p=0.90$, corresponding to the assumption that the retailer's claim is valid. 0.5 An unknown distribution has a mean of 90 and a standard deviation of 15. A more precise distance based on the normality of $\overline{X}$ is $1.960$ standard deviations, which is $ E=\frac{1.960 \sigma}{\sqrt{n}}$. A random sample of $120$ students from a large university yields mean GPA $2.71$ with sample standard deviation $0.51$. 2.4: Sampling from a Small Population - Statistics LibreTexts The probability question asks you to find a probability for the sample mean. To find the value that is two standard deviations above the expected value 90, use the formula: value = x + (#ofTSDEVs) areatotheleftofk,mean, What is the probability of finding at least one centenarian in a random sample of 200 persons from that population? We search for the area $0.9950$ in Figure $\PageIndex{5}$. Want to cite, share, or modify this book? Round the answer to the next higher value. Arrow down to A:1-PropZint. (Try other products: $(0.6)(0.4) = 0.24$; $(0.3)(0.7) = 0.21$; $(0.2)(0.8) = 0.16$ and so on). Suppose that in a population of voters in a certain region $38\%$ are in favor of particular bond issue. Since X = 90, X = 15, and n = 25. x ) A confidence interval for a population mean is an estimate of the population mean together with an indication of reliability. To use the new formula we use the line in Figure 7.1.6 that corresponds to the relevant sample size. Figure $\PageIndex{2}$ shows that when $p=0.5$ a sample of size $15$ is acceptable. x \[n = \dfrac{z^{2}\hat{p}\hat{q}}{EBP^{2}}\nonumber \], \[n = \dfrac{1.645^{2}(0.5)(0.5)}{0.03^{2}} = 751.7\nonumber \]. a. 4 years ago Figure out how many standard deviations away from the mean your proportion is, then consult a z-table and figure out the values. citation tool such as. ); The sample proportions $\hat{p}$ and $\hat{q}$ are calculated from the data: $\hat{p}$ is the estimated proportion of successes, and $\hat{q}$ is the estimated proportion of failures. Thus the population proportion $p$ is the same as the mean  of the corresponding population of zeros and ones. n lowervalueofthearea,uppervalueofthearea,mean, If 5 people are selected at random from the population, what is the probability that at least 4 of the people selected will always take medicine as prescribed? if ("frame_id"in edata) { 10% Rule of assuming "independence" between trials The mean We go to 2.0, and it was 2.02. This page titled 7.1: Large Sample Estimation of a Population Mean is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. The confidence interval can be used only if the number of successes $n\hat{p}$ and the number of failures $n\hat{q}$ are both greater than five. Thus the Central Limit Theorem applies to $\hat{p}$. The distribution is symmetric about the meanhalf the values fall below the mean and half above the mean. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo Expert Answer 100% (1 rating) Transcribed image text: If we sample 105 teachers' salaries, what is the probability that the sample mean is less than $56,500? b) If the population of grade-point averages has a normal distribution. samplesize standarddeviation Remember that the area to the right of $z_{0.05}$ is 0.05 and the area to the left of $z_{0.05}$ is 0.95. Justify your answer. ( ( Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of n. The variable n is the number of values that are averaged together, not the number of times the experiment is done. The solution is to use a different distribution, called Students $t$-distribution with $n-1$ degrees of freedom. Expert Answer The answer for the above question is Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n. An unknown distribution has a mean of 45 and a standard deviation of eight. = the mean of a sample of size 25. \[\begin{align*} P(\hat{P}\leq 0.84) &= P\left ( Z\leq \frac{0.84-\mu _{\hat{P}}}{\sigma _{\hat{P}}} \right )\\[4pt] &= P\left ( Z\leq \frac{0.84-0.90}{0.0\overline{27}} \right )\\[4pt] &= P(Z\leq -2.20)\\[4pt] &= 0.0139 \end{align*} \nonumber \]. when the level of confidence is $90\%$; when the level of confidence is $99\%$. If your target market is 29- to 35-year-olds, should you continue with your development strategy? Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. Find the probability that the sample mean is between two hours and three hours. A sample of size $49$ has sample mean $35$ and sample standard deviation $14$. A sample of size $15$ drawn from a normally distributed population has sample mean $35$ and sample standard deviation $14$. The Sample Proportion - GitHub Pages Use the calculatorprovided above to verify the following statements: When $\alpha=0.1, n=200, \hat{p}=0.43$ the EBPis $0.0577$, When $\alpha=0.05, n=100, \hat{p}=0.81$ the EBPis $0.0768$, When $\alpha=0.01, n=250, \hat{p}=0.57$ the EBPis $0.0806$. The sampling distributions are: n = 1: x 0 1 P(x) 0.5 0.5. n = 5: How do you know you are dealing with a proportion problem? Since the sample mean tends to target the population mean, we have, The central limit theorem states that for large sample sizes(, The probability that the sample mean age is more than 30 is given by. Find the probability that the sample proportion computed from a sample of size $900$ will be within $5$ percentage points of the true population proportion. Accessibility StatementFor more information contact us atinfo@libretexts.org. \[\hat{p} = \dfrac{x}{n} = \dfrac{421}{500} = 0.842\nonumber \], \[\hat{q} = 1 \hat{p} = 1 0.842 = 0.158\nonumber \], \[\alpha = 1 CL = 1 0.95 = 0.05\left(\dfrac{\alpha}{2}\right) = 0.025.\nonumber \], \[z_{\dfrac{\alpha}{2}} = z_{0.025 = 1.96}\nonumber \]. Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table. There are different formulas for a confidence interval based on the sample size and whether or not the population standard deviation is known. And so essentially we want to know the probability-- the Z-table will tell you how much area is below this value. If $\sigma$ is known: \[\bar{x}\pm z_{\alpha /2}\left ( \dfrac{\sigma }{\sqrt{n}} \right ) \nonumber \], If $\sigma$ is unknown: \[\bar{x}\pm z_{\alpha /2}\left ( \dfrac{s}{\sqrt{n}} \right ) \nonumber \]. x Solved 1) In a large population, 67 % of the people have - Chegg First we use the formulas to compute the mean and standard deviation of $\hat{p}$: To be within $5$ percentage points of the true population proportion $0.38$ means to be between $0.38-0.05=0.33$ and $0.38+0.05=0.43$. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. To learn what the sampling distribution of $\hat{p}$ is when the sample size is large. We expect that about $ (0.05)(40)=2$ of the intervals so constructed would fail to contain the population mean $\mu $, and in this simulation two of the intervals, shown in red, do. document.getElementById(frameid).contentWindow.postMessage( The true population mean falls within the range of the 95% confidence interval. 50 function loadnewq(frameid, qid) { Investors in the stock market are interested in the true proportion of stocks that go up and down each week. x Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their . } \[(\hat{p} EBP, \hat{p}+ EBP).\nonumber \]. He surveys 500 students and finds that 300 are registered voters. Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Do the results suggest that cans are filled with an amount greater than 16 ounces? The graph gives a picture of the entire situation. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. For the first question, she will pick someone else with probability 14=15. Since the population is normally distributed, the sample is small, and the population standard deviation is unknown, the formula that applies is Equation 7.2.1. so / 2 = 0.025. x \[\hat{p} =\frac{x}{n}=\frac{102}{121}=0.84\nonumber \], \[\sigma _{\hat{P}}=\sqrt{\frac{(0.90)(0.10)}{121}}=0.0\overline{27}\nonumber \], \[\left [ p-3\sigma _{\hat{P}},\, p+3\sigma _{\hat{P}} \right ]=[0.90-0.08,0.90+0.08]=[0.82,0.98]\nonumber \]. If X = _________, X = __________, and n = ___________, then XX ~ N(______, ______) by the central limit theorem for means. X is the average of both X and Using plus-four, we have $x = 13 + 2 = 15$ and $n = 50 + 4 = 54$. has a different z-score associated with it from that of the random variable X. The standard error of the mean is 4) Suppose a jar contains 17 red marbles and 17 blue marbles. If we divide the random variable, the mean, and the standard deviation by $n$, we get a normal distribution of proportions with $P $, called the estimated proportion, as the random variable. Remember that $\hat{q} = 1 \hat{p}$. The last line of that table, the one whose heading is the symbol $\infty$ for infinity and $[z]$, gives the corresponding $z$-value $z_c$ that cuts off a right tail of the same area $c$. This page titled 6.3: The Sample Proportion is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Thus the number $1.960$ in the example is $ z_{.025}$, which is $z_{\frac{\alpha }{2}}$ for $\alpha =1-0.95=0.05$. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones. d) The sample has more than 30 grade-point averages. x \[\hat{p}= 33/69 \approx 0.478\nonumber \], \[\hat{q} = 1 \hat{p} = 1 0.478 = 0.522\nonumber \]. x See answer Advertisement joaobezerra Using the binomial distribution, it is found that there is a 0.2415 = 24.15% probability that at least 4 of the people selected will always take medicine as prescribed. The summary statistics in the two samples are the same, but the $90\%$ confidence interval for the average GPA of all students at the university in "Example 4" in Section 7.1, $(2.63,2.79)$, is shorter than the $90\%$ confidence interval $(2.45,2.97)$, in "Example 6" in Section 7.1. ( Except where otherwise noted, textbooks on this site To calculate the sample size $n$, use the formula and make the substitutions. Construct a $98\%$ confidence interval for the population mean using this information. lies wholly within the interval $[0,1]$. For $95\%$ confidence the area in each tail is $\alpha /2=0.025$. ) 15 First, the underlying distribution is a binomial distribution. Solved Assume that in a large city, the distribution of the - Chegg x $\hat{p} = 0.842$ is the sample proportion; this is the point estimate of the population proportion. Construct a $90\%$ confidence interval for the mean GPA of all students at the university. The confidence intervals are constructed entirely from the sample data (or sample data and the population standard deviation, when it is known). -2- Formulas (I) Descriptive Statistics -3- xi x n 1 2 sxx x n 1i 22 1212 12 11 11 p ns n s s nn 01 y bbx 1 2 ii i x xy y b xx