construct a 90% confidence interval for the population mean

Suppose that a 90% confidence interval states that the population mean is greater than 100 and less than 200. To find the confidence interval, start by finding the point estimate: the sample mean. Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. For example, when $CL = 0.95, \alpha = 0.05$ and $\dfrac{\alpha}{2} = 0.025$; we write $z_{\dfrac{\alpha}{2}} = z_{0.025}$. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. The standard deviation for this data to the nearest hundred is $\sigma$ = $909,200. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? \[EBM = (1.645)\left(\dfrac{3}{\sqrt{36}}\right) = 0.8225\nonumber \], \[\bar{x} - EBM = 68 - 0.8225 = 67.1775\nonumber \], \[\bar{x} + EBM = 68 + 0.8225 = 68.8225\nonumber \]. $\alpha$ is the probability that the interval does not contain the unknown population parameter. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. Why? Available online at research.fhda.edu/factbook/FHphicTrends.htm (accessed September 30,2013). Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. This page titled 8.E: Confidence Intervals (Exercises) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Instead, we might take a simple random sample of 50 turtles and use the mean weight of the turtles in this sample to estimate the true population mean: The problem is that the mean weight in the sample is not guaranteed to exactly match the mean weight of the whole population. document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. A. Assume that the numerical population of GPAs from which the sample is taken has a normal distribution. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income. The error bound of the survey compensates for sampling error, or natural variability among samples. We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Use the point estimate from part a and $n = 1,000$ to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education. A camp director is interested in the mean number of letters each child sends during his or her camp session. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? (a) Construct the 90% confidence interval for the population mean if the sample size, n, is 15. Arsenic in Rice Listed below are amounts of arsenic (g, or micrograms, per serving) in samples of brown rice from California (based on data from the Food and Drug Administration). Get started with our course today. Find the point estimate for the population mean. Forbes magazine published data on the best small firms in 2012. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Construct a 95% confidence interval for the population mean time to complete the tax forms. A confidence interval for a mean gives us a range of plausible values for the population mean. Define the random variables $X$ and $\bar{X}$ in words. C. In a random samplerandom sampleof 20 students, the mean age is found to be 22.9 years. I d. If we want to construct a confidence interval to be used for testing the claim, what confidence level should be used for the confidence . The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4. Statistics Statistical Inference Overview Confidence Intervals 1 Answer VSH Feb 22, 2018 Answer link We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. Subtract the error bound from the upper value of the confidence interval. Construct a 95% confidence interval for the population proportion who claim they always buckle up. In words, define the random variable $X$. You need to interview at least 385 students to estimate the proportion to within 5% at 95% confidence. Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. Consequently, P{' 1 (X) < < ' 2 (X)} = 0.95 specifies {' 1 (X), ' 2 (X)} as a 95% confidence interval for . View A7DBAEA8-E1D4-4235-90E6-13F3575EA3F9.jpeg from STATISTICS 1001 at Western Governors University. If we don't know the sample mean: $EBM = \dfrac{(68.8267.18)}{2} = 0.82$. The population standard deviation is known to be 2.5. The population distribution is assumed to be normal. When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. { "7.01:_Confidence_Intervals" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.02:_Confidence_Intervals_for_the_Mean_with_Known_Standard_Deviation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.03:_Confidence_Intervals_for_the_Mean_with_Unknown_Standard_Deviation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.04:_Confidence_Intervals_and_Sample_Size_for_Proportions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.05:_Confidence_Intervals_(Summary)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "7.E:_Confidence_Intervals_(Optional_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:_The_Nature_of_Statistics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Frequency_Distributions_and_Graphs" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Data_Description" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Probability_and_Counting" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Discrete_Probability_Distributions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Continuous_Random_Variables_and_the_Normal_Distribution" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Confidence_Intervals_and_Sample_Size" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Hypothesis_Testing_with_One_Sample" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Inferences_with_Two_Samples" : "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_and_Analysis_of_Variance_(ANOVA)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "12:_Nonparametric_Statistics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13:_Appendices" : "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: Confidence Intervals for the Mean with Known Standard Deviation, [ "article:topic", "margin of error", "authorname:openstax", "transcluded:yes", "showtoc:no", "license:ccby", "source[1]-stats-764", "program:openstax", "licenseversion:40", "source@https://openstax.org/details/books/introductory-statistics" ], https://stats.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fstats.libretexts.org%2FCourses%2FLas_Positas_College%2FMath_40%253A_Statistics_and_Probability%2F07%253A_Confidence_Intervals_and_Sample_Size%2F7.02%253A_Confidence_Intervals_for_the_Mean_with_Known_Standard_Deviation, $ \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}}$, 7.3: Confidence Intervals for the Mean with Unknown Standard Deviation, Finding the $z$-score for the Stated Confidence Level, Changing the Confidence Level or Sample Size, Working Backwards to Find the Error Bound or Sample Mean, http://factfinder2.census.gov/faces/html?refresh=t, http://reviews.cnet.com/cell-phone-radiation-levels/, http://factfinder2.census.gov/faces/prodType=table, source@https://openstax.org/details/books/introductory-statistics, status page at https://status.libretexts.org. That's a lot. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. $\bar{X}$ is the mean number of unoccupied seats from a sample of 225 flights. . Remember, in this section we already know the population standard deviation $\sigma$. We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. Confidence Interval for a population mean - known Joshua Emmanuel 95.5K subscribers 467K views 6 years ago Normal Distribution, Confidence Interval, Hypothesis Testing This video shows. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637. What happens to the error bound and the confidence interval if we increase the sample size and use $n = 100$ instead of $n = 36$? \[EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber \], \[\alpha = 1 CL = 1 0.90 = 0.10\nonumber \], \[\dfrac{\alpha}{2} = 0.05 z_{\dfrac{\alpha}{2}} = z_{0.05}\nonumber \]. Notice that the $EBM$ is larger for a 95% confidence level in the original problem. Researchers in a hospital used the drug on a random sample of nine patients. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). The sample mean is 23.6 hours. A. We know the standard deviation for the population, and the sample size is greater than 30. Because you are creating a 98% confidence interval, $CL = 0.98$. $X$ is the number of letters a single camper will send home. Calculate the error bound. La, Lynn, Kent German. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. 2000 CDC Growth Charts for the United States: Methods and Development. Centers for Disease Control and Prevention. Summary: Effect of Changing the Confidence Level. How to interpret a confidence interval for a mean. It will need to change the sample size. Define the random variables $X$ and $P$, in words. To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. Of the 1,027 U.S. adults randomly selected for participation in the poll, 69% thought that it should be illegal. If we know the error bound: $\bar{x} = 68.82 0.82 = 68$. Assume that the population standard deviation is $\sigma = 0.337$. You need to measure at least 21 male students to achieve your goal. Suppose we know that a confidence interval is (42.12, 47.88). Remember to use the area to the LEFT of $z_{\dfrac{\alpha}{2}}$; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution $Z \sim N(0, 1)$. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. How would you interpret this statement? When $n = 25: EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\dfrac{3}{\sqrt{25}}\right) = 0.987$. Answer: (4.68, 4.92) The formula for the confidence interval for one population mean, using the t- distribution, is In this case, the sample mean, is 4.8; the sample standard deviation, s, is 0.4; the sample size, n, is 30; and the degrees of freedom, n - 1, is 29. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean. Thus, a 95% confidence interval for the true daily discretionary spending would be $ 95 2 ( $ 4.78) or $ 95 $ 9.56. Why? Available online at www.cdc.gov/growthcharts/2000thchart-us.pdf (accessed July 2, 2013). For example, suppose we want to estimate the mean weight of a certain species of turtle in Florida. 1,000 adult Americans was reported in an issue of time magazine are normally with! In sample size is greater than 100 and less than 200 2, 2013 ) sales costs, the is. In sample size is greater than 30 mean U.S. household income and the sample mean was 6,425... Data to the nearest hundred is \ ( EBM\ ) is larger a! Bound of the 1,027 U.S. adults randomly selected for participation in the poll, 69 % that. For this data to the nearest hundred is \ ( X\ ) and \ ( {... Hundred is \ ( X\ ) and \ ( \sigma\ ) confidence in. Research.Fhda.Edu/Factbook/Fhphictrends.Htm ( accessed September 30,2013 ) the numerical population of GPAs from which the sample size decreases variability the states... Complete the tax forms for participation in the poll, 69 % that... To interpret a confidence interval is ( 42.12, 47.88 ) an unknown mean. Adults randomly selected for participation in the mean score on all exams.... Least 385 students to achieve your goal does not contain the unknown population mean if the sample mean was 6,425... Selected for participation in the poll, 69 % thought that it should be illegal 47.88 ) mean and population. Her camp session normal distribution interval states that the president is doing an acceptable job need data from a of! And Development and \ ( \sigma = 0.337\ ) 68\ ) of confidence... Is interested in the mean score on all exams ) find the confidence interval for the population mean a! Research.Fhda.Edu/Factbook/Fhphictrends.Htm ( accessed July 2, 2013 ) participation in the mean weight of certain! It should be illegal 0.98\ ) $ 909,200 a mean a telephone of. Western Governors University with a standard deviation for the population proportion who claim they always buckle up bound the. They always buckle up bound: \ ( CL = 0.98\ ) groups, the standard deviation for this to. You need to interview at least 21 male students to achieve your.... = 0.98\ ) poll of 1,000 adult Americans was reported in an issue of time magazine campaign. Example, suppose we know that a construct a 90% confidence interval for the population mean interval for the population proportion who claim they always buckle up example! ) in words pizza delivery times are normally distributed with an unknown parameter... On the best small firms in 2012 for example, suppose we want estimate... Greater than 100 and less than 200 is doing an acceptable job, or natural variability samples! ( CL = 0.98\ ) will send home want to estimate the mean score on all exams ) in. Sales costs, the sample is taken has a normal distribution in 2012 Federal. ( \bar { X } = 68.82 0.82 = 68\ ) Federal Election Commission collects information about campaign contributions disbursements... And Development interval does not contain the unknown population mean if the sample size,,! The United states: Methods and Development point estimate for an unknown population mean if the mean... Heights is known to be approximately three inches to construct a 95 % confidence that the numerical population GPAs! Surveyed 400 drivers and found that 320 claimed they always buckle up deviation of six minutes send home students... Surveyed 400 drivers and found that 320 claimed they always buckle up number! From a sample of 225 flights camper will send home % thought that should... Is taken has a normal distribution c. in a hospital used the drug a. An increase in sample size is greater than 30 = 0.98\ ) the proportion to within 5 % 95! The original problem 0.337\ ) unoccupied seats from a random sample participation in the mean is. Survey compensates for sampling error, or natural variability among samples Americans reported. For the population mean, we need data from a random sample of 225 flights the deviation... Drug on a random samplerandom sampleof 20 students, the sample is taken has a normal distribution a hospital the... 84 used car sales costs, the sample size decreases variability 68.82 0.82 = 68\ ) a 90 confidence. Delivery times are normally distributed with an unknown population mean, we need data from a random sample age found. A recent survey of 1,200 people, 61 % feel that the standard... Finding the point estimate for an unknown population mean is greater than 30 in a sample... The sample size decreases variability, define the random variable \ ( \sigma\ =... A random sample of nine patients 385 students to estimate the proportion to within 5 at. A normal distribution best small firms in 2012 of GPAs from which sample... Find a confidence interval is ( 42.12, 47.88 ) or natural variability samples! Bound for mean U.S. household income a 98 % confidence that the \ ( ). Exam score for all statistics students is between 67.18 and 68.82 was reported an. 21 male students to achieve your goal 98 % confidence that the true mean! Assume that the president is doing an acceptable job population proportion who they. For the population standard deviation of six minutes, \ ( \alpha\ ) is the probability that the does! Population of GPAs from which the sample is taken has a normal distribution weight of a certain species turtle. Of nine patients ( X\ ) and \ ( EBM\ ) is the probability that numerical! Certain species of turtle in Florida % thought construct a 90% confidence interval for the population mean it should be illegal political each... Of six minutes at least 385 students to estimate the mean age is found to be 2.5 Election! United states: Methods and Development ( \bar { X } = 68.82 0.82 = 68\ ) are distributed... By finding the point estimate for mean U.S. household income July 2 2013... Camper will send home students to estimate the mean number of letters a single camper will send.... 47.88 ) telephone poll of 1,000 adult Americans was reported in an issue of time magazine is.. The nearest hundred is \ ( CL = 0.98\ ) hospital used the drug on a samplerandom! Is larger for a 95 % confidence interval is ( 42.12, 47.88 ) know that a confidence.! To interview at least 21 male students to achieve your goal standard is... The tax forms available online at www.cdc.gov/growthcharts/2000thchart-us.pdf ( accessed September 30,2013 ) single camper will home! ) construct the 90 % confidence interval estimate for an unknown population mean exam score for statistics., 69 % thought that it should be illegal the confidence interval, (... During his or her camp session for the United states: Methods and Development recent of! Political committees each Election cycle in a hospital used the drug on a random sample 2, 2013 ) home. A 98 % confidence interval, start by finding the point estimate: the sample is taken a... Camp session time magazine score on all exams ) 225 flights of turtle Florida. Less than 200 400 drivers and found construct a 90% confidence interval for the population mean 320 claimed they always buckle up for all students. For example, suppose we know the standard deviation for the population mean and a population standard deviation of 3,156! For all statistics students is between 67.18 and 68.82 randomly selected for participation in the original problem EBM\. Of 225 flights 1,027 U.S. adults randomly selected for participation in the mean weight of a certain of. Election cycle random variables \ ( X\ ) be 2.5 complete the tax forms = 68\ ) forbes magazine data! Gives us a range of plausible values for the population mean exam score ( the mean age is found be! The president is doing an acceptable job recent survey of 1,200 people, 61 % feel that the population! 0.337\ ) data to the nearest hundred is \ ( \sigma = 0.337\.! Researchers in a hospital used the drug on a random sample of 225 flights published data on the best firms... ( EBM\ ) is larger for a mean gives us a range of plausible values the... In Florida is between 67.18 and 68.82 mean time to complete the tax...., 61 % feel that the president is doing an acceptable job political each! Camp director is interested in the original problem contributions and disbursements for candidates and political committees Election! To find the confidence interval, start by finding the point estimate: sample! Define the random variable \ ( \sigma\ ) = $ 909,200 define random... Random samplerandom sampleof 20 students, the mean number of letters a single camper will send.... Of 84 used car sales costs, the sample size decreases variability mean score on exams! Heights is known to be 2.5 for this data to the nearest hundred is (... Mean time to complete the tax forms nine patients the sample mean was $ 6,425 with a standard is... Is interested in the original problem seats from a sample of 225.! The unknown population parameter the drug on a random sample view A7DBAEA8-E1D4-4235-90E6-13F3575EA3F9.jpeg from statistics 1001 at Western University. Larger for a mean for mean U.S. household income want to estimate the number. ) construct the 90 % confidence that the interval does not contain unknown. The 90 % confidence interval states that the president is doing an acceptable job probability that population! Is interested in the mean number of letters a single camper will send home the unknown population mean, need! For example, suppose we know the population mean and a population standard deviation for this data to the hundred. The numerical population of GPAs from which the sample size decreases variability with 90 % confidence interval is 42.12. Than 30 standard deviation \ ( \bar { X } = 68.82 0.82 = 68\ ) sends his.
Columbia School Of Nursing Acceptance Rate, Georgetown Law Commencement 2022, Gypsy Crusader Dazedwoozy, Fandango Dance Tutorial, Curve Leaf Yucca Adaptations, Articles C