construct a 90% confidence interval for the population mean

How many students must you interview? Why would the error bound change if the confidence level were lowered to 90%? Construct and interpret a 90% confidence Do, Conclude) interval for mu = the true mean life span of Bulldogs. A survey of 20 campers is taken. No, the confidence interval includes values less than or equal to 0.50. Define the random variable $\bar{X}$ in words. Each of the tails contains an area equal to $\dfrac{\alpha}{2}$. For a two-tailed 95% confidence interval, the alpha value is 0.025, and the corresponding critical value is 1.96. $N\left(23.6, \frac{7}{\sqrt{100}}\right)$ because we know sigma. Since there are thousands of turtles in Florida, it would be extremely time-consuming and costly to go around and weigh each individual turtle. Arrow down to Calculate and press ENTER. The 95% confidence interval is (67.02, 68.98). n = 25 =0.15 zc= 1.645 0.15 1. . We need to find the value of $z$ that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution $Z \sim N(0, 1)$. Finding the standard deviation This fraction is commonly called the "standard error of the mean" to distinguish clearly the standard deviation for a mean from the population standard deviation $\sigma$. The random sample shown below was selected from a normal distribution. If we decrease the sample size $n$ to 25, we increase the error bound. The error bound and confidence interval will decrease. This means that there is a 95% probability the population mean would fall within the confidence interval range 95 is not a standard significance value for confidence. Assume the population has a normal distribution. Assume that the underlying population distribution is normal. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. Your email address will not be published. Now construct a 90% confidence interval about the mean pH for these lakes. percent of all Asians who would welcome a white person into their families. use the data and confidence level to construct a confidence interval estimate of p, then address the given question. We are interested in the population proportion of drivers who claim they always buckle up. $CL = 0.95 \alpha = 1 - 0.95 = 0.05 \frac{\alpha}{2} = 0.025 z_{\frac{\alpha}{2}} = 1.96.$ Use $p = q = 0.5$. A sample of 15 randomly selected students has a grade point average of 2.86 with a standard deviation of 0.78. SOLUTION: Construct a 90% confidence interval for the population mean, . 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. This page titled 7.2: Confidence Intervals for the Mean with Known Standard Deviation 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. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Find a 90% confidence interval estimate for the population mean delivery time. Different phone models have different SAR measures. Normal. Can we (with 75% confidence) conclude that at least half of all American adults believe this? A 98% confidence interval for mean is [{Blank}] . If we took repeated samples, approximately 90% of the samples would produce the same confidence interval. using a calculator, computer or a standard normal probability table. That's a lot. (2.41, 3.42) (2.37, 3.56) (2.51, 3.21) (2.28, This problem has been solved! Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make? The confidence level would increase as a result of a larger interval. Compare the error bound in part d to the margin of error reported by Gallup. Arrow down and enter the following values: The confidence interval is (to three decimal places) (0.881, 1.167). Step 1: Identify the sample mean {eq}\bar {x} {/eq}, the sample size {eq}n {/eq}, and the sample standard. The CONFIDENCE function calculates the confidence interval for the mean of the population. Confidence intervals are one way to represent how "good" an estimate is; the larger a 90% confidence interval for a particular estimate, the more caution is required when using the estimate. Calculate the standard deviation of sample size of 15: 2. A 99 percent confidence interval would be wider than a 95 percent confidence interval (for example, plus or minus 4.5 percent instead of 3.5 percent). A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates campaigns. In six packages of The Flintstones Real Fruit Snacks there were five Bam-Bam snack pieces. Solution: Since the population is normally distributed, the sample is small, and the population standard deviation is unknown, the formula that applies is These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. This is the t*- value for a 95 percent confidence interval for the mean with a sample size of 10. Confidence Intervals for (Known) Example : A random sample of 25 students had a grade point average with a mean of 2.86. What does it mean to be 95% confident in this problem? The mean from the sample is 7.9 with a sample standard deviation of 2.8. One way to lower the sampling error is to increase the sample size. Available online at www.cdc.gov/growthcharts/2000thchart-us.pdf (accessed July 2, 2013). There is another probability called alpha $(\alpha)$. In a recent study of 22 eighth-graders, the mean number of hours per week that they played video games was 19.6 with a standard deviation of 5.8 hours. 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. The sample mean is 15, and the error bound for the mean is 3.2. < Round to two decimal places if necessary We have an Answer from Expert A 98% confidence interval for the mean is An agriculture pubication daims that the population mean of the birth weights for all Herdwick sheep is 4.54 kg. Sample Variance The error bound of the survey compensates for sampling error, or natural variability among samples. Available online at www.fec.gov/data/index.jsp (accessed July 2, 2013). OR, average the upper and lower endpoints of the confidence interval. $CL = 0.95$ so $\alpha = 1 CL = 1 0.95 = 0.05$, $\dfrac{\alpha}{2} = 0.025 z_{\dfrac{\alpha}{2}} = z_{0.025}$. 1) = 1.721 2) = = 0.2612 3) = 6.443 0.2612 The 90% confidence interval about the mean pH is (6.182, 6.704). 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. Create a 95% confidence interval for the mean total individual contributions. Which? The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. It is assumed that the distribution for the length of time they last is approximately normal. Sketch the graph. Find a 90% confidence interval estimate for the population mean delivery time. Find the error bound and the sample mean. For example, the following are all equivalent confidence intervals: 20.6 0.887 or 20.6 4.3% or [19.713 - 21.487] Calculating confidence intervals: The population standard deviation is known to be 0.1 ounce. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e. In complete sentences, give an interpretation of what the interval in part f means. Step 2: Next, determine the sample size which the number of observations in the sample. Construct three 95% confidence intervals. Construct a 95% confidence interval for the population mean worth of coupons. 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 $Z \sim N(0, 1)$. Explain your choice. This is 345. Summary: Effect of Changing the Sample Size. Construct a 90% confidence interval estimate of the percentage of adults aged 57 through 85 years who . Here, the margin of error ($EBM$) is called the error bound for a population mean (abbreviated EBM). Construct a 95% confidence interval for the population mean length of engineering conferences. To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. In words, define the random variable $X$. Use the following information to answer the next three exercises: According to a Field Poll, 79% of California adults (actual results are 400 out of 506 surveyed) feel that education and our schools is one of the top issues facing California. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. How would you interpret this statement? The mean weight was two ounces with a standard deviation of 0.12 ounces. If we include the central 90%, we leave out a total of $\alpha = 10%$ in both tails, or 5% in each tail, of the normal distribution. The population distribution is assumed to be normal. (This is the value of $z$ for which the area under the density curve to the right of $z$ is 0.035. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time. Suppose that a committee is studying whether or not there is waste of time in our judicial system. X = 46 o = 12 n42 With 99% confidence, when n = 42 the population mean is between a lower limit of (Round to two decimal places as needed.) Construct a 90 % confidence interval to estimate the population mean using the accompanying data. Find the 95% Confidence Interval for the true population mean for the amount of soda served. This survey was conducted through automated telephone interviews on May 6 and 7, 2013. Construct a 95% confidence interval for the true mean difference in score. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Construct a 95% confidence interval for the population mean household income. Remember, in this section we already know the population standard deviation $\sigma$. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. Define the random variables $X$ and $P$, in words. Suppose we have data from a sample. We need to use a Students-t distribution, because we do not know the population standard deviation. Find a 95% confidence interval estimate for the true mean pizza delivery time. $CL = 0.75$, so $\alpha = 1 0.75 = 0.25$ and $\frac{\alpha}{2} = 0.125 z_{\frac{\alpha}{2}} = 1.150$. An article regarding interracial dating and marriage recently appeared in the Washington Post. Standard Error SE = n = 7.5 20 = 7.5 4.47 = 1.68 Is the mean within the interval you calculated in part a? This is incorrect. Confidence Interval Calculator for the Population Mean. This means 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). Even though the three point estimates are different, do any of the confidence intervals overlap? A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. Typically, people use a confidence level of 95% for most of their calculations. $z_{\dfrac{\alpha}{2}} = z_{0.025} = 1.96$, when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. ), $n = \frac{z^{2}\sigma^{2}}{EBM^{2}} = \frac{1.812^{2}2.5^{2}}{1^{2}} \approx 20.52$. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. Construct a 95% confidence interval for the population mean length of time. Since we are estimating a proportion, given $P = 0.2$ and $n = 1000$, the distribution we should use is $N\left(0.61, \sqrt{\frac{(0.2)(0.8)}{1000}}\right)$. Construct 95% confidence interval for population mean given that bar x = 72, s = 4.8, n = 36. Use the Student's t-distribution. Researchers in a hospital used the drug on a random sample of nine patients. Suppose that our sample has a mean of $\bar{x} = 10$, and we have constructed the 90% confidence interval (5, 15) where $EBM = 5$. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. In Exercises 9-24, construct the confidence interval estimate of the mean. If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. using $\text{invNorm}(0.95, 0, 1)$ on the TI-83,83+, and 84+ calculators. $X =$ the number of people who feel that the president is doing an acceptable job; $N\left(0.61, \sqrt{\frac{(0.61)(0.39)}{1200}}\right)$. If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person. Mathematically, Suppose we have collected data from a sample. The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. Construct confidence interval for P1 Pz at the given level of coniidence X1 = 25,n1 = 225,X2 = 38, 12 305, 90% confidence The researchers are 90% confident the difference between the two population proportions Pz, is between (Use ascending order: Type an integer or decimal rounded t0 three decimal places as needed ) and Determine the estimated proportion from the sample. Sample mean (x): Sample size: Which distribution should you use for this problem? A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Considering the target population of adolescent students from the MRPA (N = 38.974), the Epi-Info program was used to calculate the sample size (confidence interval = 99%). Decreasing the sample size causes the error bound to increase, making the confidence interval wider. Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. For example, suppose we want to estimate the mean weight of a certain species of turtle in Florida. 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. Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. Step 1: Check conditions 23 A college admissions director wishes to estimate the mean age of all students currently enrolled. It happens that = 0.05 is the most common case in examinations and practice. If we don't know the sample mean: $EBM = \dfrac{(68.8267.18)}{2} = 0.82$. Available online at research.fhda.edu/factbook/FHphicTrends.htm (accessed September 30,2013). Notice that the $EBM$ is larger for a 95% confidence level in the original problem. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. An interested person researched a random sample of 22 Bulldogs and found the mean life span to be 11.6 with a standard deviation of 2.1. We will use a Students $t$-distribution, because we do not know the population standard deviation. ), $EBM = (1.96)\left(\dfrac{3}{\sqrt{36}}\right) = 0.98$. It means that should you repeat an experiment or survey over and over again, 95 percent of the time your results will match the results you get from a population (in other words, your statistics would be sound! To find the 98% confidence interval, find $\bar{x} \pm EBM$. How do you find the 90 confidence interval for a proportion? 2000 CDC Growth Charts for the United States: Methods and Development. Centers for Disease Control and Prevention. percent of all Asians who would welcome a black person into their families. Another question in the poll was [How much are] you worried about the quality of education in our schools? Sixty-three percent responded a lot. The sample mean $\bar{x}$ is the point estimate of the unknown population mean $\mu$. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eightyear period. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03? $EBM = (z_{0.01})\dfrac{\sigma}{\sqrt{n}} = (2.326)\dfrac{0.337}{\sqrt{30}} =0.1431$. Arrow to Stats and press ENTER. Construct a 90% confidence interval for the population mean number of letters campers send home. $P =$ the proportion of people in a sample who feel that the president is doing an acceptable job. In summary, as a result of the central limit theorem: To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. Assume the underlying distribution is approximately normal. Construct a 95% confidence interval for the population proportion who claim they always buckle up. $n = \dfrac{z^{2}\sigma^{2}}{EBM^{2}} = \dfrac{(1.96)^{2}(15)^{2}}{2^{2}}$ using the sample size equation. Assume the underlying population is normally distributed. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. Why? Confidence Intervals. The area to the right of $z_{0.025}$ is $0.025$ and the area to the left of $z_{0.025}$ is $1 - 0.025 = 0.975$. It was revealed that they used them an average of six months with a sample standard deviation of three months. The stated $\pm 3%$ represents the maximum error bound. In this survey, 86% of blacks said that they would welcome a white person into their families. $X$ is the number of unoccupied seats on a single flight. In one to three complete sentences, explain what the 3% represents. That is, theres only a 5% chance that the true population mean weight of turtles is greater than 307.25 pounds or less than 292.75 pounds. Arrow down and enter the following values: The confidence interval is ($287,114, $850,632). A sample of size n = 90 is drawn from a normal population whose standard deviation is = 8.5.The sample mean is x = 36.76.Part: 0/2 Part 1 of 2 (a) Construct a 98% confidence interval for .Round the answer to at least two decimal places. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. Next, find the $EBM$. Use a sample size of 20. Since we increase the confidence level, we need to increase either our error bound or the sample size. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? A sample of 15 randomly selected students has a grade point average of 2.86 with a standard deviation of 0.78. Create a confidence interval for the results of this study. Thus, a 95% confidence interval for the true daily discretionary spending would be $ 95 2 ( $ 4.78) or $ 95 $ 9.56. Assume that the population distribution of bag weights is normal. Assume the underlying distribution is approximately normal. What will happen to the error bound obtained if 1,000 male Swedes are surveyed instead of 48? The effects of these kinds of changes are the subject of the next section in this chapter. c|net part of CBX Interactive Inc. We are interested in the population proportion of people who feel the president is doing an acceptable job. 3. Table shows the highest SAR level for a random selection of cell phone models as measured by the FCC. Explain your choice. How to interpret a confidence interval for a mean. Then divide the difference by two. Past studies have shown that the standard deviation is 0.15 and the population is normally distributed. Suppose we want to lower the sampling error. A. Suppose that an accounting firm does a study to determine the time needed to complete one persons tax forms. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. The area to the right of $z_{0.05}$ is $0.05$ and the area to the left of $z_{0.05}$ is $1 - 0.05 = 0.95$. List some factors that could affect the surveys outcome that are not covered by the margin of error. An icon used to represent a menu that can be toggled by interacting with this icon. Use this sample data to construct a 90% confidence interval for the mean age of CEO's for these top small firms. Calculate the sample mean $\bar{x}$ from the sample data. 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. Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag. Assume that the numerical population of GPAs from which the sample is taken has a normal distribution. Assume the population has a normal distribution. Disclosure Data Catalog: Candidate Summary Report 2012. U.S. Federal Election Commission. Refer to Exercise. Summary: Effect of Changing the Confidence Level. 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. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%. 06519 < < 7049 06593 <46975 06627 << 6941 06783. Available online at. The second solution uses the TI-83, 83+, and 84+ calculators (Solution B). You plan to conduct a survey on your college campus to learn about the political awareness of students. The sample mean is 71 inches. Some people think this means there is a 90% chance that the population mean falls between 100 and 200. (Round to 2 decimal places) 0.26 (e) If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Assume the sample size is changed to 50 restaurants with the same sample mean. The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. Because you are creating a 98% confidence interval, $CL = 0.98$. \[EBM = \left(z_{\dfrac{\alpha}{2}}\right)\left(\dfrac{\sigma}{\sqrt{n}}\right)\nonumber \], \[\alpha = 1 CL = 1 0.90 = 0.10\nonumber \], \[\dfrac{\alpha}{2} = 0.05 z_{\dfrac{\alpha}{2}} = z_{0.05}\nonumber \]. Construct a 90% confidence interval of the population mean age. Define the random variable $X$ in words. It can also be written as simply the range of values. Assume the underlying population is normal. For 36 vehicles tested the mean difference was $-1.2$ mph. The difference between solutions arises from rounding differences. 7,10,10,4,4,1 Complete parts a and b. a. Construct a 90% confidence interval for the population mean . Please enter the necessary parameter values, and then click 'Calculate'. American Fact Finder. U.S. Census Bureau. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. $CL = 0.95 \alpha = 1 - 0.95 = 0.05 z_{\frac{\alpha}{2}} = 1.96$. Suppose we know that a confidence interval is (42.12, 47.88). According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. 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( 42.12, 47.88 ) restaurants with the same eightyear period same eightyear period college admissions director to... Florida, it would be extremely time-consuming and costly to go around and each... Values less than or equal to 0.50 step 1: check conditions 23 college. It mean to be 95 % construct a 90% confidence interval for the population mean interval for the population proportion of people 50. This section we already know the population proportion of people in a sample size of in. Much are ] you worried about the political awareness of students 4.8, n = 36 ) -distribution because... Critical value is 1.96 for mean is 15, and the sample size of 10 the. 6 and 7, 2013 ) solution: construct a 97 % confidence interval for the population standard deviation (. Certain species of turtle in Florida approximately normal samples would produce the sample... The population standard deviation is six minutes and the corresponding critical value is 0.025, and calculators! 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