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71 48 14.8 Cell Styles ~ 10.09 16.26 7 56 18. 79 Styles 7 Next two questions are related to the f lated to the following sample data M118 128 You gather a random sample of court records from the files of adults charged with assault to 129 obtain an interval estimate of the mean age of those charged. Yo charged. You obtain the following B D G 130 sample of age data. 131 40 1245 John obtained in his random sample 132 33 43 1265 133 29 27 29 23 23 1476 134 1045 135 71 The confidence interval with a 95%% confidence level is 22.1 42.6 137 b 25.0 39.6 138 0 27.1 37.6 139 d 28 6 36 1 1280 1060 140 1280 1150 141 12 You want to obtain a m obtain a margin of error of 2 years for your 95%% interval estimate. What is 142 the minimum sample size to reach this margin of error? Use the standard deviation from ! 48 49 1 Question 1-6 are based on John's sample data. 143 your previous sample as the planning value. The point estimate of the mean monthly rental charge is._. 144 145 6 28 b 1213.74 35 1.167.06 146 0 1.122.17 147 d 55 148 2 The sample variance is. 19 Next two questions are related to the following sample data 29.618.06 b 150 A random sample of employees of a multinational corporation yielded the following annual 151 salary data (in dollars). 152 128550 66420 97390 66710 66470 101320 19340 63240 153 53540 65830 85670 85690 84580 44580 90900 3 154 61500 The estimated variance of the sample mean is. 94780 82040 76040 133890 88910 8090 63910 83020 ####8# 63 b 28,182.25 155 88540 84450 58520 79020 102690 83090 107650 103390 156 90690 72170 68540 102040 96570 99210 88130 67620 65 d 296.66 66 157 90440 90900 83420 81920 92950 97510 103780 57760 67 4 158 71670 77550 112910 81080 111330 82970 93330 113250 68 34.19 The margin of error for a 95% interval estimate for the population mean rental charge is. 159 86760 71960 63510 126730 10340 103660 77840 62830 69 29.73 160 102540 71480 67060 104650 74660 75110 77020 70520! 70 25.86 2248 95900 62610 76310 64640 41530 86600 73890 80000 162 76940 116470 78870 83770 93190 78660 ' 107050 110040 73 5 he 30%. interval estimate for the population mean rental charge is. 163 97360 72450 99860 121570 75460 89280 60720 136220! 74 1,207.58 70340 116380 85860 85890 56880 62950 84300 75 1,217.84 1.317.00 120100 76 1,228.10 1,296.48 165 90910 58890 80540 101910 127830 93660 97700 93950! d 1.242.75 1,281.83 89320 95450 84230 107240 92440 92360 67660 80100 103780 119110 81720 92190 92920 115060 87180 82810 In the previous question. to obtain a margin of error of :$25 for a 95% interval 116090 77240 117030 90630 86430 96760 66980! estimate the minimum sample size is For planning value, use the 168 77120 standard deviation from John's sample data. 169 91150 112250 68330 90260 81540 83090 117250 98410 170 70090 86100 89940 120590 58990 86840 86590 110040 197 171 89220 69050 95870 70240 90090 87630. 58990 124730 176 172 75280 88130 78150 115340 127660 82690 82570 59320 64890 86 173 70650 55970 118200 136250 96120 94710 74710 In interval estimation of a population parameter, increasing the sample size would - 82210 122520 64260 91400 123790 110050 39380 81260 increase the margin of sampling error, making the interval estimate more precise. 174 b reduce the margin of sampling error, making the interval estimate more saga 175 68590 63570 127970 76560 118910 100310 86150 75710 reduce the margin of san margin of sampling error, increasing the confidence level. 176 94890 82530 101960 92910 87580 84670 86850 98840 91 d reduce the margin of sampling error, diminishing the error probability. 177 24650 109770 104910 99910 80890 59740 90060 71840 92 178 59700 72410 101560 77810 107520 118290 112860 101320 In interval estimation of a population parameter, increasing a would: 73110 9500 reduce the margin of samplin error, and reduce the confidence level 82640 73810 75230 64630 68920 89500 bre reduce the margin of sampling error, and reduce the error probability. 179 60540 94650 55380 63100 reduce the margin of sampling etto 75620 134290 sampling error, and make the and make the interval estimate less p 180 66560 97910 898 increase the margin of sampling er and make the interval estimate more precise. 181 85810 42250 100560 88310 91960 106460 69790 78380 182 69650 45860 72220 57320 92950 47350 95110 99 90070 00 J Next TWO ques ons are related to the following sample data: 183 73820 76330 98670 98290 68240 54240 95090 $3750 its to build a confidence interval with a 95%% confidence 91490 70900 level for the mean 101200 I students. oh s. She selects a random 30340 84280 91620 84770 108290 02 sample of students and obtains the following commuting distance data. The data show 92700 13 the distance from the student residence to the campus in miles). 185 89070 44820 95910 74710 81240 77810 99620 16.5 186 66540 68060 125980 64540 54560 144040 91750 39830 13.2 9.2 15.8 187 55120 114910 89210 80280 98570 106000 74200 66690! The point estimate of the mean salary is, $94,656 $91,899 $89,222 $86,623 8988288988 The standard error of the mean is, 14 The margin of error for a 95%% interval estimate of the population mean salary is $2,523 $2,293 MADE WITH The interval estimate 2,085 $1.895 O Photo Editor
