Confidence Interval Sample Problems

Problem Summary 1- 5: Difference between two means 6-7: Proportions 8-11: Single sample estimation

Problem 1 (Diff btwn Two Means) A taxi company is trying to decide: whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are Brand A: x1 — 36,300 kilometers, s1 = 5,000 kilometers. Brand B: x2 = 38,100 kilometers, s2 = 6,100 kilometers. Compute a 95% confidence interval for ma — mb assuming the populations to be approximately normally distributed. Assume variances are not equal. Explanation: Estimating difference ma — mb, assume variances are not equal, also s1 and s2 means the variance is unknown. Given those conditions we use: where

Problem 1 Solution Because we are looking for a 95% confidence interval, a = 0.05 and a/2 = 0.025, so we go to t-table and find the value that corresponds to v = 21, and a = 0.025. That value is 2.080

Problem 2 (Diff btwn Two Means) Two kinds of thread are being compared for strength. Fifty pieces of each type of thread are tested under similar conditions. Brand A had an average: tensile strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand B had an average tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct a 95% confidence interval for the difference of the population means. Explanation: Question asks for difference in population means, gives actual standard deviation (not sample standard deviation) this means s12 and s22 are known, also n>30 so n is considered large. Because variance or standard deviation is known, and because n is large, we use this equation: If standard deviation was unknown, but n was large, you should still use this equation.

Problem 2 Solution Because we are looking for a 95% confidence interval, a = 0.05 and a/2 = 0.025, so we go to z-table and find the value that corresponds to an area of 0.025 under the curve. That corresponds to z = 1.96

Problem 3 (Diff btwn Two Means) In a batch chemical process, two catalysts are being compared for their effect on the output of the process reaction. A sample of 12 batches was prepared using catalyst 1 and a sample of 10 batches was obtained using catalyst 2. The 12 batches for which catalyst 1 was used gave an average yield of 85 with a sample standard deviation of 4, and the second sample gave an average of 81 and a sample standard deviation of 5. Find a 90% confidence interval for the difference between the population means, assuming that the: populations are approximately normally distributed with equal variances. Explanation: Estimating difference ma — mb, assume variances are equal, also because a SAMPLE standard deviation is given that means the variance is unknown. Given those conditions we use: where

Problem 3 Solution This is the square root of Sp2 Degrees of freedom = v = n1+n2-2

Problem 4 (Diff btwn Two Means) Students may choose between a 3-semester-hour course in physics without, labs and a 4-semester- hour course with labs. The final written examination is the same for each section. The results are shown below: 3-semester-hour course 4-semester-hour course n1=12 n2=18 x1 = 84 x2 = 77 s1 = 4 s2 = 6 Find a 99%. confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances. Explanation: Estimating difference ma — mb, assume variances are equal, also because a SAMPLE standard deviation is given that means the variance is unknown. Given those conditions we use: where

Problem 4 Solution This is the square root of Sp2 Degrees of freedom = v = n1+n2-2

Problem 5 (Diff btwn Two Means) The following data represent the running times of films produced by two motion-picture companies. Compute a 90% confidence interval for the difference between the average running times of films produced by the two companies. Assume that the running-time differences are approximately normally distributed with unequal variances. Explanation: Estimating difference ma — mb, assume variances are not equal, also you are given raw sample data and have to calculate sI and sII , the variance is unknown. Given those conditions we use: where

Problem 5 Solution You are given raw sample data so you must calculate xI sI xII and sII. xI = (103+94+110+87+98)/5 = 98.4 xII = (97+82+123+92+175+88+118)/7 = 110.7 Use this equation to calculate the variance: sI = 8.375 sII = 32.185

Problem 6 (Proportion) Compute a 98% confidence interval for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives. Explanation: You are asked to compute a confidence interval for the proportion of defects. So use the following:  

Problem 7 (Proportion) A manufacturer of compact disk players uses a set of comprehensive tests to access the electrical function of its product. All compact disk players must pass all tests prior to being sold. A random sample of 500 disk players resulted in 15 failing one or more tests. Find a 90% confidence interval for the proportion of compact disk players from the population that will pass the test. Explanation: You are asked to compute a confidence interval for the proportion disk that will PASS. So use the following: Read carefully!!!! You are asked to create an interval for disk that will PASS but you are giving failing disk info. So if 15 out of 500 failed, 485 out of 500 passed

Problem 8 (Single Estimation) An electrical firm manufactures light bulbs that have a length of life that, is approximately normally distributed with a standard deviation of 40 hours. A sample of 30 bulbs and has an average life of 780 hours Find a 96% confidence interval for the population mean of all bulbs produced by this firm. Explanation: You are asked to compute a confidence interval using one sample of light bulbs. You use the single estimation equation for s known because the actual standard deviation is given. Also n is equal to exactly 30 so you can use the following equation because one or both of the conditions (s known and n is large) is met.

Problem 8

Problem 9 (Single Estimation) Many cardiac patients wear implanted pacemakers to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch. Explanation: You are asked to compute a confidence interval for the mean of connector modules. You use the single estimation equation for s known because the actual standard deviation is given. Also n larger than 30 so you can use the following equation because one or both of the conditions (s known and n is large) is met.

Problem 9

Problem 10 (Single Estimation) A machine is producing metal pieces that are cylindrical in shape. A sample of 7 pieces is taken with an average diameter of 1.0056 and a sample standard deviation of 0.0245. Find a 99% confidence interval for the mean diameter of pieces from this machine, assuming an approximate normal distribution. Explanation: You are asked to compute a confidence interval for the mean the piece diameter. You use the single estimation equation for s unknown because a SAMPLE standard deviation is given. Also n is small (less than 30). Because s is unknown AND n is small you use: Degrees of freedom = v = n-1

Problem 10 Solution Degrees of freedom = v = n-1

Problem 11 (Single Estimation) A random sample of 12 shearing pins is taken in a study of the Rockwell hardness of the head on the pin. Measurements on the Rockwell hardness were made for each of the 12, yielding an average value of 48.50 with a sample standard deviation of 1.5. Assuming the measurements to be normally distributed, construct a 90% confidence interval for the mean Rockwell hardness. Explanation: You are asked to compute a confidence interval for the mean the piece diameter. You use the single estimation equation for s unknown because a SAMPLE standard deviation is given. Also n is small (less than 30). Because s is unknown AND n is small you use: Degrees of freedom = v = n-1

Problem 11 Solution Degrees of freedom = v = n-1