# estimating population proportion

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confidence? Did you have an idea for improving this content? Clopper-Pearson is used by stats::binom.test() in R as well. He surveys 500 students and finds that 300 are registered voters. Remember that the plus-four method assume an additional four trials: two successes and two failures. This can swamp any pettifogging details about t versus z. Suppose a baseball team manager takes a random sample of 120 pitches from the rival pitcher. Press ENTER.Arrow down to and enter 421.Arrow down to and enter 500.Arrow down to C-Level and enter .95.Arrow down to Calculate and press ENTER.The confidence interval is (0.81003, 0.87397). That page gives you all the details of the method, with worked-out The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B). Narrower confidence intervals give more precise interval estimates for the population proportion, but this is true only if the intervals contain the population proportion. There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Although you can’t do the 104 successes in sample, 11037−104 = 10933 For the normal distribution of proportions, the z-score formula is as follows. Since CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 (α) = 0.025. For the sample of size 100, the confidence interval is, $0.55\text{}±\text{}1.96\sqrt{\frac{(0.55)(0.45)}{100}}\text{}=\text{}0.55\text{}±\text{}0.098\text{}=\text{}(0.452,0.648)$, For the sample of size 400, the confidence interval is, $0.55\text{}±\text{}1.96\sqrt{\frac{(0.55)(0.45)}{400}}\text{}=\text{}0.55\text{}±\text{}0.049\text{}=\text{}(0.501,0.599)$. If X is a binomial random variable, then X ~ B(n, p) where nis the number of trials and p is the probability of a success. $\displaystyle{X}~{N}{({n}{p},\sqrt{{{n}{p}{q}}})}$, If we divide the random variable, the mean, and the standard deviation by sample size must increase by the, Always write an interpretation of your CI. there’s one more source of un-preciseness that neither method temperature of 98.6°? Arrow down to A:1-PropZint. For example, using a 95% confidence level, we are 95% confident that the population proportion falls within the interval. This is way more precision than we can really justify, but I just want you to see that the five results are all (slightly) different. see how long they would operate a wireless mouse. endobj X is binomial. this, because the program is a lot easier. Of the 500 people surveyed, 421 responded yes – they own cell phones. Available online at http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx (accessed July 2, 2013). The new sample size, then, is n + 4, and the new count of successes is x + 2. large must your sample be? “Teens, Social Media, and Privacy.” PewInternet, 2013. If you don’t see how to start a The interpretation of a confidence interval depends on the confidence level. 12 0 obj The standard error of that sample proportion will be, $$SE(p) = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.708(1-0.708)}{1737}} = 0.011$$, And our 95% confidence interval (so that we’ll use $$\alpha$$ = 0.05) for the true population proportion, $$\pi$$, of EVD cases with definitive outcomes, who will die is << /S /GoTo /D (section.1.4) >> Lesson 20 Classwork (lower bound, upper bound)= (p’ – EBP, p’ + EBP) = (p’ – $\displaystyle({z}_{\frac{{\alpha}}{{2}}})(\sqrt{\frac{{p'q'}}{{n}}})$, p’+ $\displaystyle({z}_{\frac{{\alpha}}{{2}}})(\sqrt{\frac{{p'q'}}{{n}}})$) Calculate , and proceed to find the confidence interval. For a proportion, the appropriate standard deviation is $\displaystyle\sqrt{\frac{{pq}}{{n}}}$. (0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%. The margin of error is related to the confidence level. endobj The random variable P′(read “P prime”) is that proportion, $\displaystyle{P'}=\frac{{X}}{{n}}$, (Sometimes the random variable is denoted as $\displaystyle\hat{P}$, read “P hat”.). The confidence interval is (0.564, 0.636). Because we are using the “plus-four” method, we will use x = 6 + 2 = 8 and n = 25 + 4 = 29. p’ = $\displaystyle\frac{{x}}{{n}} =\frac{{8}}{{29}}$ = 0.276. Caution! Please submit your feedback or enquiries via our Feedback page. So larger sample sizes are the preferred way to decrease the error and create narrower confidence intervals. Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1). Sample Size Calculator for Estimating a Single Proportion . Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013). (Estimating population proportion, chapter 8) Assume that pitching a fastball (event E1) and pitching a breaking ball (event E2) are mutually exclusive and collectively exhaustive events. It is the range in which the true population proportion is estimated to be and is often expressed in percentage points (e.g., ±2%). We have seen that 395 / 462 subjects (or a proportion of 0.855) fall in the “normal range” in our sample. The confidence interval is (0.355, 0.602). you want a margin of error no more than 3.5% at 95% confidence, how Remember that the area to the right of z0.05 is 0.05 and the area to the left of z0.05 is 0.95. successive approximations done by MATH200A. endobj p’=  $\displaystyle\frac{{x}}{{n}} =\frac{{421}}{{500}}$ = 0.842. p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. It is easy to see that the plus-four method has the greatest impact on smaller samples. If there is less variability in the sampling distribution, the standard error is smaller. (Confidence intervals) endobj The other elements of the tidied result are shown below. problem and check your answer with the step-by-step explanations. q′ = 1 – p′ The variable p′ has a binomial distribution that can be approximated with … 24 0 obj For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table. But we are less confident that 90% confidence intervals contain the population proportion. If $\displaystyle{P'}{\sim}{N}$(p, $\displaystyle\sqrt{\frac{{pq}}{{n}}}$) then the z-score formula is z = $\displaystyle\frac{{p'-p}}{{\sqrt{pqn}}}$. examples. Recall that in the long run, 10% of these intervals will not contain the population proportion at all! We welcome your feedback, comments and questions about this site or page. xڥXY�E~�_я�T���eØ1�����a� ���I��4����:�j����\$���p�V��z����.-lR��ݯ�Z���*)L��3�n:N�ZWw���W�7���*�����2JVAJa���ջ���D��8Q���Lu�0|����=<>N�Ƙ���6�� 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. (Confidence Interval Belt Graphs) As of September 14, 2014, a total of 4,507 confirmed and probable cases of Ebola virus disease (EVD) had been reported from West Africa. << /S /GoTo /D (chapter.1) >> “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. ... Our intuition tells us that larger samples should give more precise estimates of the population proportion. The Score approach is also used by stats::prop.test() and creates CIs by inverting p-values from score tests. 36 0 obj Available online at http://www.zogbyanalytics.com/news/299-americans-neither-worried-nor-prepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). The first solution is step-by-step (Solution A). The sample proportion will not be exactly equal to the population proportion, but values of the sample proportion from random samples tend to cluster around the actual value of the population proportion.