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Confidence Intervals for the Population Mean, Using the t Distribution A sample of taxicabs had the following consumption of gasoline, in gallons, on a particular day: 32, 16, 25, 22, 27, 19, 20. Assume that the distribution of gasoline usage is normal. A. Find the appropriate t values for 90% and 95% confidence intervals. B. Find the mean gasoline usage and the standard deviation of gasoline use. C. Find the sample standard error. D. Calculate the 90 and 95% confidence intervals for the mean daily use of gasoline by all taxicabs. Solutions: A. Since n = 7, degrees of freedom = 7 – 1 = 6. See below to find t values of 1.943 and 2.447. B. The most appropriate way to find these numbers is with a statistical calculator. You should find that ⎯x = 23 and s = 5.416026. _ _ C. The sample standard error s⎯x = s/√n = 5.416026/√7 = 2.047065 D. The confidence interval is given by ⎯x ± t × s⎯x ⇒ for the 90% confidence interval 23 ± 1.943 × 2.0470650 or 23 ± 3.98. We are 90% confident that the average daily gasoline use in the population of taxicabs is between 19.020and 26.98 gallons, or between 19 and 27 gallons. For the 95% confidence interval, we have 23 ± 2.447 × 2.0470650 or 23 ± 5. We are 95% confident that the mean daily gasoline use for all taxicabs is at least 18 but not more than 28 gallons. Student's t distribution Confidence Level This column is used to locate the degrees of freedom 0.800 0.900 0.950 0.980 Significance Level for Two-Tailed Test 0.200 0.100 0.050 0.020 Significance Level for One-Tailed Test 0.990 df 0.005 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 0.100 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 0.050 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 0.025 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.201 2.179 2.160 2.145 2.131 2.120 2.110 2.101 0.010 31.821 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 0.010 This row is the confidence levels expressed as decimal fractions. 63.657 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 A. df = n – 1 = 7 – 1 = 6. Find the confidence level in the top row of labels: 0.900. 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 A. For the 95% confidence interval, use the column headed 0.950. A sample of 57 adults passed through a federally funded job training program. After sixteen weeks, the 57 were found and their current weekly income was recorded. In the sample, the mean weekly income was $412 with standard deviation $53. Calculate a 98% confidence interval for the mean wage of all workers who have participated in this particular training program. Solution: _ The confidence _interval is ⎯x ± t × s⎯x where s⎯x = s/√n. In this problem, ⎯x = 412, s = 53 and n = 57. Thus, s⎯x = 53/√57 = 53/7.549834435 = 7.02. For a 98% confidence interval, the t value = 2.395 (see below). Thus the confidence interval is 412 ± 2.395 × 7.02 ⇒ 412 ± 16.81. With 98% confidence, we can say the mean wage of all participants in this program is between $395.19 and $428.81. Confidence Level 0.800 0.900 0.950 0.980 0.990 Significance Level for Two-Tailed Test 0.200 0.100 0.050 0.020 0.010 Significance Level for One-Tailed Test df 47 48 49 50 51 52 53 54 55 56 57 58 59 60 0.100 1.300 1.299 1.299 1.299 1.298 1.298 1.298 1.297 1.297 1.297 1.297 1.296 1.296 1.296 0.050 1.678 1.677 1.677 1.676 1.675 1.675 1.674 1.674 1.673 1.673 1.672 1.672 1.671 1.671 0.025 2.012 2.011 2.010 2.009 2.008 2.007 2.006 2.005 2.004 2.003 2.002 2.002 2.001 2.000 0.010 2.408 2.407 2.405 2.403 2.402 2.400 2.399 2.397 2.396 2.395 2.394 2.392 2.391 2.390 0.005 2.685 2.682 2.680 2.678 2.676 2.674 2.672 2.670 2.668 2.667 2.665 2.663 2.662 2.660 With n = 57, there are n – 1 = 56 degrees of freedom. Use the column headed 0.980 for a 98% confidence interval.

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