Discuss the relationship between the margin of error and the standard error of the mean. For a (1 -)-level confidence interval for a population mean when the population standard deviation is known, the margin of error equals the standard error of the meanDiscuss the relationship between the margin of error and the standard error of the mean. For a (1 -)-level confidence interval for a population mean when the population standard deviation is known, the margin of error equals the standard error of the mean population standard deviation. subtracted from multiplied by divided by added toDiscuss the relationship between the margin of error and the standard error of the mean. For a (1-()-level confidence interval for a population mean when the population standard deviation is known, the margin of error equals the standard error of the mean population standard deviation. half of the twice the theDiscuss the relationship between the margin of error and the standard error of the mean. For a (1 -)-level confidence interval for a population mean when the population standard deviation is known, the margin of error equals the standard error of the mean z-score that has area 2a to its right. population standard deviation. z-score that has area a to its right. a z-score that has area , to its right. sample size. sample mean.Each year, a professor obtains estimates for the mean age, u, of all millionaires in a certain country. Suppose that one year's study involved a simple random sample of 37 millionaires in the country whose mean was 59.78 years with a sample standard deviation of 12.78 years. Complete parts (a) and (b) below. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. If, for next year's study, a confidence interval for u is to have a margin of error of 3 years and a confidence level of 90%, determine the required sample size. (Round up to the nearest whole number.) b. Why did you use the sample standard deviation, s = 12.78, in place of the population standard deviation o in your solution in part (a)? Why is it permissible to do so? The sample standard deviation was used in place of o because It is permissible to do so becauseEach year, a professor obtains estimates for the mean age, u, of all millionaires in a certain country. Suppose that one year's study involved a simple random sample of 37 millionaires in the country whose mean was 59.78 years with a sample standard deviation of 12.78 years. Complete parts (a) and (b) be Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. - X Areas under the standard normal curve, 2 of 2 X Areas under the standard normal curve, 1 of 2 nine th Areas under the Second decimal place in z art (a)1 standard normal curve Second decimal place in z 0.00 0.01 0.02 0.03 0.04 0.03 0.06 0.07 0.08 0.09 Areas under the standard normal curve 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53: 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57: 0.0000 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.614 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 03 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.651 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 04 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.687 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 05 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.727 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.754 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78: 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.813 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.838 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.862 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88: 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.901 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.0026 13 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.917 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 14 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.931 0.0036 0.0037 0.0038 0.0039 0.0040 0.0041 0.0043 0.0044 0.0045 0.0047 0.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.0062 15 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.944 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.954 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.0082 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.963 0.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.0107 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.970 0.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.0139 19 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.976 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.0228 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.981 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.985 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.0287 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.989 0.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.0359 23 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.991 0.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 0.0436 0.0446 24 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99: 0.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.0548 0.0559 0.0571 0.0582 0.0594 0.0606 0.0618 0.0630 0.0643 0.0655 0.0668 -1. 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.995 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.996 0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.0808 -14 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.997 0.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 28 0 9974 0 9975 09976 ( 0977 0 0977 ( 0978 0 9979 0 0979 0 9980 n qq5Each year, a professor obtains estimates for the mean age, u, of all millionaires in a certain country. Suppose that one year's study involved a simple random sample of 37 millionaires in the country whose mean was 59.78 years with a sample standard deviation of 12.78 years. Complete parts (a) and (b) below. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. If, for next year's study, a confidence interval for u is to have a margin of error of 3 years and a confidence level of 90%, determine the required sample size. 50 (Round up to the nearest whole number.) b. Why did you use the sample standard deviation, s = 12.78, in place of the population standard deviation o in your solution in part (a)? Why is it permissible to do so? The sample standard deviation was used in place of o because It is permissible to do so because the prior study's mean is large enough to make s more accurate than o. the prior study's mean is small enough to make s more accurate than o. when available, a sample standard deviation is always more appropriate than o. the prior study's sample size is small enough to make s more accurate than o. is unknown. the prior study's sample size is large enough to make s more accurate than o.Each year, a professor obtains estimates for the mean age, u, of all millionaires in a certain country. Suppose that one year's study involved a simple random sample of 37 millionaires in the country whose mean was 59.78 years with a sample standard deviation of 12.78 years. Complete parts (a) and (b) below. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. If, for next year's study, a confidence interval for u is to have a margin of error of 3 years and a confidence level of 90%, determine the required sample size. 50 (Round up to the nearest whole number.) b. Why did you use the sample standard deviation, s = 12.78, in place of the population standard deviation o in your solution in part (a)? Why is it permissible to do so? The sample standard deviation was used in place of o because It is permissible to do so because the sample size for the prior study is small enough. the prior study's margin of error is greater than the desired margin of error. the mean for the prior study is small enough. the prior study's margin of error is close enough to the desired margin of error. the sample size for the prior study is large enough. the prior study's margin of error is less than the desired margin of error. the mean for the prior study is large enough.In a study, the mean number of days that 132 adolescents in substance abuse treatment used medical marijuana in the last 8 months was 108.58. Assuming the population standard deviation is 31 days, a 95% confidence interval for the mean number of days, u, of medical marijuana use in the last 8 months of all adolescents in substance abuse treatment is from 103.29 days to 113.87 days; this interval's margin of error is 5.29 days. Complete parts (a) through (d) below. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. The mean number of days that 33 adolescents in substance abuse treatment used medical marijuana in the last 8 months was 106.23. Find a 95% confidence interval for u based on that data. The 95% confidence interval is from ] day(s) to ] day(s). (Round to two decimal places as needed.) b. Compare the 95% confidence intervals obtained in part (a) and in the original study by drawing a graph. Choose the correct graph below. O A. O B. 95% Cl for 95% Cl for (n = 132) (n= 132) 95% Cl for P 95% Cl for (n = 33) 106 120 100 108 116 O C. OD 95% Cl for 95% Cl for (n = 132) (n= 132) 95% Cl for p 95% Cl for p (n = 33) (n = 33) 108 108 120 c. Compare the margins of error for the two 95% confidence intervals. The margin of error for the interval obtained in part (a) is ] day(s), which is the margin of error of the interval obtained in the original study. Round to two decimal places as needed.) d. What principle is being illustrated? The principle being illustrated here is that increasing the the margin of error of a confidence intervalIn a study, the mean number of days that 132 adolescents in substance abuse treatment used medical marijuana in the last 8 months was 108.58. Assuming the population standard deviation is 31 days. a 95% confidence interval for the mean number of days. I. of medical marijuana use in the last 8 months of all adolescent in substance abuse treatment is from 103.29 days to 113.87 days; this interval's margin of error is 5.29 days. Complete parts (a) through (d) below. Areas under the standard normal curve, 1 of 2 Areas under the standard normal curve, 2 of 2 X as 106.23. Find a 95% confidence interval for Areas under the Second decimal place in z Areas under the Second decimal place in z standard normal curve 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 standard normal curve 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.0 0.0000 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.535 graph below. 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57: 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61- OB. 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 03 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.651 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 04 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.687 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.722 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 05 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.754 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78: 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.812 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.838 O D. 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 UNNU ! 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.862 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.0026 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88: 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.901 0.0036 0.0037 0.0038 0.0039 0.0040 0.0041 0.0043 0.0044 0.0045 0.0047 13 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.917 0.0048 0.0049 0.0051 0.0052 0.0054 0.0055 0.0057 0.0059 0.0060 0.0062 UN NN 14 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.931 0.0064 0.0066 0.0068 0.0069 0.0071 0.0073 0.0075 0.0078 0.0080 0.0082 15 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94 0.0084 0.0087 0.0089 0.0091 0.0094 0.0096 0.0099 0.0102 0.0104 0.0107 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.954 0.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.0139 1.6 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0. 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.970 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.0228 19 obtained in the original study. 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.976 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.0287 -1! 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.981 0.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.0359 -11 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.985 0.0367 0.0375 0.0384 0.0392 0.0401 0.0409 0.0418 0.0427 0.0436 0.0446 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.989 0.0455 0.0465 0.0475 0.0485 0.0495 0.0505 0.0516 0.0526 0.0537 0.0548 23 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99 0.0559 0.0571 0.0582 0.0594 0.0606 0.0618 0.0630 0.0643 0.0655 0.0668 -12 24 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99: of a confidence interval 0.0681 0.0694 0.0708 0.0721 0.0735 0.0749 0.0764 0.0778 0.0793 0.0808 25 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99: 0.0823 0.0838 0.0853 0.0869 0.0885 0.0901 0.0918 0.0934 0.0951 0.0968 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.996 0.0985 0.1003 0.1020 0.1038 0.1056 0.1075 0.1093 0.1112 0.1131 0.1151 2.6 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.997 0.1170 0.1190 0.1210 0.1230 0.1251 0.1271 0.1292 0.1314 0.1335 0.1357 28 0 9974 0 9975 09976 0 0977 0 0977 0 0978 0 0979 0 0979 0 0980 n qqsIn a study, the mean number of days that 132 adolescents in substance abuse treatment used medical marijuana in the last 8 months was 108.58. Assuming the population standard deviation is 31 days, a 95% confidence interval for the mean number of days, u, of medical marijuana use in the last 8 months of all adolescents in substance abuse treatment is from 103.29 days to 113.87 days; this interval's margin of error is 5.29 days. Complete parts (a) through (d) below. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. a. The mean number of days that 33 adolescents in substance abuse treatment used medical marijuana in the last 8 months was 106.23. Find a 95% confidence interval for u based on that data. The 95% confidence interval is from 95.65 day(s) to 116.81 day(s). (Round to two decimal places as needed.) b. Compare the 95% confidence intervals obtained in part (a) and in the original study by drawing a graph. Choose the correct graph below. OA. Q B. 95% Cl for 95% Cl for (n = 132) (n = 132) 35% Callfor u 96% Cl for (n = 33) C 92 106 190 100 108 116 O C. OD 95% Cl for 95% Cl for (n = 132) (n= 132) 95% Cl for 95% Cl for (n = 33) 108 116 c. Compare the margins of error for the two 95% confidence intervals for a fixed confidence level. The margin of error for the interval obtained in part (a) is 1.58 day(s), which is greater than the margin of error of the interval obtained in the original regardless of changes to any other values used in computing the interval. (Round to two decimal places as needed.) d. What principle is being illustrated? for a fixed sample size. The principle being illustrated here is that increasing the sample size decreases the margin of error of a confidence interval