Confidence Intervals - Lecture Notes | MATH 203, Exams of Statistics

Download Confidence Intervals - Lecture Notes | MATH 203 and more Exams Statistics in PDF only on Docsity! Dr. Neal, Spring 2009 MATH 203 Confidence Intervals Confidence Interval for the Mean Let x1 , x2 , . . . , xn be a random sample of a measurement X on a population Ω with unknown mean and (unknown) standard deviation . Let x be the sample mean and let S be the sample deviation. We wish to use x and S to approximate , within a certain margin of error and with a certain “level of confidence.” Initially, we shall assume that we are sampling from a “large” population. Recall: (i) For a normally distributed measurement X ~ N( , ) , the bounds that contain inner probability r = 1 − are given by ± z /2 , where z /2 is the z -score for the desired confidence-level probability which is usually 0.90, 0.95, 0.98, or 0.99. These z -scores are 1.645, 1.96, 2.326, and 2.576. (ii) For large samples, the distribution of all possible sample means x is approximately N , n       . (It is exactly N , n       for any sample size when the measurements are known to be normally distributed.) Thus, the bounds that contain the various x with probability r = 1 − are given by ± z /2 n . That is, with probability r = 1 − , a sample mean x is within ± z /2 n of . Therefore is also within ± z /2 n of x with probability r . These bounds give us a “confidence interval” for estimating : ≈ x ± z /2 n Arbitrary Measurements: Need large samples. Normal Measurements: Any sample size works. However, this formula requires that we know the true population standard deviation , which is often unknown. But for large samples, we may replace with an estimate, which may be the sample deviation S or an upper bound U . Thus, ≈ x ± z /2 S n or ≈ x ± z /2 U n Arbitrary measurements with large samples Dr. Neal, Spring 2009 Example 1. A nationwide survey of first-year college students Ω asked “About how many minutes X do you study on a typical weeknight?” The mean response of 900 students was x = 97.4 minutes with S = 25.8 minutes. Give a 99% confidence interval for the mean study time of all first-year students. Explain the interval in words. Solution. (By Hand) With the large sample of size n = 900, we can assume that S = 25.8 is a reasonable approximation of the standard deviation . The 0.99 z -score is 2.576; thus, ≈ x ± z /2 S n = 97.4 ± 2.576 × 25.8 900 = 97.4 ± 2.2. That is, 95.2 ≤ ≤ 99.6. (By TI-83/84) Press STAT, scroll right to TESTS, and press 7 for ZInterval. Set the Inpt by highlighting Stats and pressing ENTER. Next, enter the values of = 25.8, x = 97.4, n = 900, and C-Level = .99. Then scroll down to Calculate and press ENTER. STAT TESTS 7 Enter Stats Output In words: The average study time per night of all first-year college students is somewhere from 95.2 to 99.6 minutes. Example 2. Below are the SAT scores for a sample of students from one school district. (Here, Ω = all students taking the test in this district; X = SAT score.) The scores are known to be normally distributed with = 130. 1270 850 1030 1300 1060 1160 1220 1280 1020 1070 1040 1050 1160 1000 1400 1040 990 1130 Give a 95% confidence interval for the mean SAT score from that district. Solution. Enter the data into list L1. Bring up the ZInterval screen from the STAT TESTS menu. Set the Inpt by highlighting Data and pressing ENTER. Adjust the value of , the list, frequency, and confidence level. Then press ENTER on Calculate. The TI will compute the statistics from the data in the list, then display the confidence interval (1054.9, 1175.1) and summary statistics. (Note: Because of the normally distributed scores, we do not need a large sample here.) ≈ 1115 ± 1.96 ×130 18 = 1115 ± 60.06 or 1054.94 ≤ ≤ 1175.06. The mean SAT score for students in this district is between 1054.94 and 1175.06. Dr. Neal, Spring 2009 Then we solve for n to obtain n ≥ z /2 / e( )2 . But when is unknown, we cannot simply replace with S because we are trying to determine the required sample size before taking the sample. In that case, we should use an upper bound U . So the required sample size needed to obtain a margin of error of no more than e with level of confidence r = 1 − is given by n ≥ z /2 e       2 n ≥ z /2 U e       2 where U = d − c 2 If is known If is unknown Note: n is always rounded up to the nearest integer. Example 5. A random sample of 1000 students gained an average of v x = 22 points on the SAT mathematics exam when retaking the test. The change in score among all retakes in the past has been normally distributed with a standard deviation of = 50. (a) Give a 95% confidence interval for the mean change in score among the population Ω of all students retaking the test. (b) What sample size is needed to estimate the mean change in Math SAT score to within ± 2 points with 95% confidence? Solution. (a) ≈ x ± z /2 n = 22 ± 1.96 ×50 1000 = 22 ± 3.099 . (b) We simply apply the sample size formula n ≥ (z /2 / e) 2 with z /2 = 1.96, = 50, and e = 2 to find that we would need a sample of size of at least 2401 to obtain a margin of error of ± 2 with 95% confidence. (A smaller sample might suffice, but a sample of size 2401 guarantees the result.) Example 6. We wish to estimate the average length of newborn babies. From years of observation, we may assume that the lengths range from 8 inches to 24 inches (excluding the rare extreme cases). (a) If we wish to estimate within 0.75 inches with 95% confidence, then what sample size is required? (b) If we wish to estimate within 0.5 inches with 98% confidence, then what sample size is required? (c) A random sample of 900 newborns gave a mean length of x = 19.2 inches. Find a 95% confidence interval for the true mean length. Solution. (a) Here U = (24 – 8)/2 = 8. For 95% confidence, z /2 ≈ 1.96; thus, we choose n ≥ z /2 U e       2 = 1.96 ×8 0.75       2 ≈ 437.09. Hence, we need a sample size of at least n = 438. Dr. Neal, Spring 2009 (b) Now we need n ≥ 2.326 × 8 0.5       2 ≈ 1385.03, or a sample size of at least 1386. (c) Using U = 8 as upper bound for the , we can say, ≈ x ± z /2 U n = 19.2 ± 1.96 × 8 900 = 19.2 ±0.523 inches. (Using U gives the largest possible margin of error for a sample of size 900 and a 95% confidence level.) Sample Size for Small Populations Suppose now that the population under study has known finite size N . Then a confidence interval for is given by ≈ x ± z /2 n N − n N −1 . If we want the error to be no more than e , then we can set z /2 n N − n N −1 ≤ e and solve for n . However, we first must replace by its upper bound U = (d − c) / 2 . Then solving for n , we obtain n ≥ N z /2 U e       2 N − 1+ z /2 U e       2 where U = d − c 2 or n ≥ N z /2 e       2 N − 1+ z /2 e       2 if is known Example 7. Suppose there are 900 applicants to the business college that requires a minimum GPA of 2.5. We want to find a 95% confidence interval for the mean GPA of applicants that has no more than a 0.05 margin of error. Find the required random sample size to be chosen from the 900 applicants to obtains such a confidence interval. Solution. First, an upper bound for is U = (d − c) / 2 = (4 – 2.5)/2 = 0.75. Now using a z -score of 1.96 for 95% confidence, a margin of error of e = 0.05, and a population size of N = 900, we must choose a sample size n at least as large as 900 1.96 × 0.75 0.05       2 899 + 1.96 × 0.75 0.05       2 = 900 29.4( )2 899 + 29.4( )2 = 441.16. So we require a random sample of size n = 442 from the 900 applicants. Note: For an extremely “large population” the required sample size would be 1.96 × 0.75 0.05       2 = 29.42 , which rounds up to 865. Dr. Neal, Spring 2009 Practice Exercises 1. Suppose we want a maximum margin of error of 20 when finding a 99% confidence interval for the average GRE of Graduate School applicants, where a minimum score of 1200 out of a possible 1600 is required. Find the necessary random sample size for such a confidence interval in the following cases: (a) Among a huge pool of applicants nationwide. (b) Among a select pool of 500 applicants. (c) From the select pool of 500, a random sample of size 160 gives a sample mean of 1380.4. Give a 95% confidence interval for the true mean in this case. 2. Here are the daily intakes of calcium (in mg) for a group of women between the ages of 18 and 24 who agreed to participate in a study. 808 882 1062 970 909 802 374 416 784 997 651 716 438 1420 1425 948 1050 976 572 403 626 774 1253 549 1325 446 465 1269 671 696 1156 684 1933 748 1203 2433 1255 1100 Find a 95% confidence interval for the average intake of calcium for all similar women between the ages of 18 and 24. (Use S as a reasonable estimate of .) 3. A sample of WKU 90's alumni were asked to give the number of years needed to graduate. The results were: Years 3 3.5 4 4.5 5 5.5 6 6.5 7 Responses 2 5 12 14 16 10 8 3 1 (a) Find a 98% confidence interval for the average number of years needed to graduate for all 90's alumni. (b) Excluding the rare extreme cases, all graduates took from 3 to 8 years to graduate. If we wanted to estimate the average number of years needed with a maximum margin of error of 0.25 with 98% confidence, what sample size would be needed? (c) Suppose the sample is just from the 350 chemistry majors. Now find a 98% confidence interval for the average number of years needed to graduate for all 90's chemistry majors. (d) If the range is still 3 to 8 years, what sample size of the 350 chemistry majors would be needed to estimate the average number of years needed with a maximum margin of error of 0.25 with 98% confidence?