(a) A random sample of 200 voters in a town is selected, and 111 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit. (b) What can be asserted with 96% confidence about the possible size of the error if the fraction of voters favoring the annexation suit is estimated to be 0.62? Click here to view page 1 of the normal probability table. Click here to view page 2 of the normal probability table. (a) The confidence interval is ☐
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- Respiratory Rate Researchers have found that the 95 th percentile the value at which 95% of the data are at or below for respiratory rates in breath per minute during the first 3 years of infancy are given by y=101.82411-0.0125995x+0.00013401x2 for awake infants and y=101.72858-0.0139928x+0.00017646x2 for sleeping infants, where x is the age in months. Source: Pediatrics. a. What is the domain for each function? b. For each respiratory rate, is the rate decreasing or increasing over the first 3 years of life? Hint: Is the graph of the quadratic in the exponent opening upward or downward? Where is the vertex? c. Verify your answer to part b using a graphing calculator. d. For a 1- year-old infant in the 95 th percentile, how much higher is the walking respiratory rate then the sleeping respiratory rate? e. f.0.0 .5000 5040 5080 .5120 5160 .5199 .5239 5279 .5319 .5359 0.1 .5398 5438 5478 .5517 .5557 5596 .5636 5675 5714 .5753 0.2 .5793 5832 5871 .5910 5948 5987 .6026 .6064 .6103 6141 0.3 .6179 6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 6879 0.5 .6915 .6950 .6985 .7019 .7054 7088 7123 .7157 .7190 7224 0.6 .7257 7291 .7324 .7357 .7389 .7422 7454 .7486 .7517 7549 0.7 7580 7611 7642 7673 .7704 7734 .7764 .7794 7852 7823 0.8 .7881 815 .7910 .7939 7967 .7995 .8023 8051 .8078 .8 106 8133 0.9 8186 .8212 8238 8264 8289 8315 .8340 8365 8389 1.0 8413 8438 .8461 8485 8508 .8531 8554 .8577 8599 8621 1.1 8643 8665 .8686 8708 .8729 8749 8770 8962 .8790 8810 8830 1.2 8849 8869 .8888 8907 .8925 .8944 .8980 8997 .9015 1.3 9032 .9049 .9066 .9082 .9099 9115 .9131 .9147 9162 .9177 1.4 9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 9306 .9319 1.5 .9332 .9345 9357 .9370 .9382 9394 .9406 .9418 9429 9441 1.6 .9452 .9463 9474 .9484 .9495 9505 .9515…Q.2: Mr. Ahmed is planning to recruit some additional sales staff for his company Al Khalif LLC. Since the sale varies from time to time, he decided to recruit staff if average sales is more than RO 35000. Following are the details of his company's sales for past 50 days. Sales 00- 10-20 20-30 30-40 40-50 50-60 60-70. 70- ('000RO) 10 80 No. of 05 07 10 05 08 05 05 05 days a) What would you suggest Mr. Ahmed regarding recruitment of sales staff? Give reason for your answer. b) What is the mode sale of the company? Why do you calculate mode sales?
- Jassessments/427594/variants/61/090/ e/5/ 1 of 1: May Field Trips post on Monday, Apri. DISMISS 150 152 154 156 158 160 162 164 166 Minimum: smallest amount Maximum: largest amount Q1: median of lower half Q2: median of ALL the data Q3: median of upper half minimum a. 156 b. 153 91 C. 158 Q2 d. 155Find P(Z>-2.46) .00 .01 02 .03 .04 Los .06 .07 L08 .09 0.0 0.1 .0000 .0040 .0438 .0080 .0120 .0160 .0199 .0239 .0398 .0478 .0279 .о319 .0359 .0517 .0557 .0596 0636 .0675 1064 0.2 .0793 .0832 .0871 .0714 .0753 .0910 .1293 .0948 .0987 1026 0.3 .1179 1217 .1255 .1103 .1141 .1331 .1368 .1406 1443 .1480 0.4 1554 .1591 .1628 1664 .1700 .1517 0.5 .1915 .1736 1772 1808 .1844 1879 .1950 .1985 .2019 .2054 2088 .2123 0.6 2257 2291 .2324 2157 2190 .2224 .2357 .2389 2422 .2454 .2486 0.7 .2580 2611 .2642 .2517 .2549 .2673 .2704 .2734 .2764 .2794 2823 0.8 .2881 2910 .2939 2967 .2852 .2995 .3023 .3051 .3078 .3106 .3133 0.9 .3159 3186 .3212 .3238 .3264 .3289 3315 3340 3365 3389 1.0 .3413 .3438 .3461 .3485 .3508 .3531 3554 .3577 .3599 .3621 1.1 .3643 3665 .3686 3708 .3729 .3749 .3770 .3790 .3810 .3830 1.2 3849 .3869 .3888 .3907 .3925 3944 .3962 .3980 .3997 .4015 1.3 4032 4049 .4066 .4082 4099 4115 4131 .4147 .4162 .4177 1.4 4192 .4207 .4222 .4236 .4251 .4265 4279 .4292 .4306 .4319 1.5 4332 4345 .4357…Evaluate the probability when n = 9 and p = 0.1. (Round your answer to three decimal places.) P(xs 1) n USE SALT You may need to use the appropriate appendix table to answer this question.
- 14. Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.054 Click to view page 1 of the table.1 Click to view page 2 of the table.² The cumulative area corresponds to the z-score of (Round to three decimal places as needed.)Period People (Y) Ocupation percentage (X) 2009 1T 46,977,904 59.7 2009 2T 47,453,163 59.3 2009 3T 48,738,589 60.2 2009 4T 48,903,792 60.4 2010 1T 48,069,274 59.9 2010 2T 49,133,132 60.2 2010 3T 49,190,032 58.7 2010 4T 48,478,718 58.8 2011 1T 48,505,168 59.3 2011 2T 49,482,112 59.5 2011 3T 50,127,032 59.7 2011 4T 50,772,496 60.5 2012 1T 50,192,842 60.3 2012 2T 51,477,178 60.7 2012 3T 51,927,050 60.7 2012 4T 51,317,999 60 2013 1T 50,847,242 60 2013 2T 51,895,865 60.4 2013 3T 52,034,353 60.2 2013 4T 52,370,886 60.7 2014 1T 51,559,018 60.3 2014 2T 51,836,752 59.8 2014 3T 52,192,043 59.6 2014 4T 52,108,400 59.4 2015 1T 52,007,842 59.7 2015 2T 52,623,721 59.6 2015 3T 53,179,919 59.7 2015 4T 53,809,017 60.3 2016 1T 52,918,649 59.7 2016 2T 53,539,565 59.6 2016 3T 54,226,803 59.9 2016 4T 54,034,800 59.6 2017 1T 53,681,720 59.7 2017 2T 54,068,791 59.3 2017 3T 54,369,915 59.1 2017 4T 54,696,638 59.3 2018 1T 54,590,773…z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003-3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005-3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007-3.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010-2.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014-2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019-2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026-2.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036-2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188…
- The Harris Corporation University of Florida study to determine whether a manufacturing process performed at a remote location can be established locally. Test devices (pilots) were set up at both the old and new locations and voltage readings on 30 production runs at each location were obtained. The data are given in the table attached. a. Compare the mean voltage readings at the two locations using a 95% confidence interval. b. Based on the interval, part a, does it appear that the manufacturing process can be established locally.Find P(Z>-1.46) 00 01 .02 .03 .04 05 .06 .07 08 .09 0.0 .0000 0040 .o080 .0120 .0160 0199 .0239 .0279 .0319 .0359 0.1 .0398 0438 0478 .0517 .0557 .0596 .06 36 0675 .0714 .0753 0.2 .0793 OB 32 08/1 .0910 .0948 .0987 .1026 .1064 .1103 .1141 LEEL" .1368 .1736 0.3 .11/9 .1217 .1255 1293 .1406 .1143 1480 -1517 0.4 .1554 .1591 1628 .1GG4 .1700 .1772 .1808 .1844 .1879 0.5 .1915 1950 1985 .2019 .2054 .2088 .2123 2157 2190 2224 0.6 O.6 2257 2291 2324 2357 2389 2422 2454 2486 2517 2549 0.7 2580 2611 2642 2673 2704 2134 2764 2194 2823 2852 : 0.8 2881 2910 2939 2967 2995 3023 3051 30/8 3106 3133 0.9 3159 3186 3212 3238 3264 8289 3340 3365 SLEE 1.0 3413 8438 8461 3485 3508 3531 3554 3577 3599 3621 1.1 3643 3665 3686 3708 3729 3749 3770 3790 3810 38 30 1.2 3849 3869 3888 3907 3925 3944 3980 3997 4015 296E 1.3 4032 1049 4066 4082 4099 4115 1131 4147 4162 11// 1.4 4192 1207 1236 1251 4265 1306 1319 1.5 4332 4357 4370 A I82 4394 440G 4418 4429 4441 1.6 4452 4463 4474 4484 4495 4505 4515 4525 4535 4545…It was hypothesized that there will be a difference in average test score in chemistry class between high schoolers who watched Breaking Bad (G1) and high schoolers who did not watch Breaking Bad (G2). Calculate and use the appropriate statistical test to analyze the below. Alpha criteria is set at 0.01. Watched Breaking Bad Did Not Watch Breaking Bad 86 99 90 79 91 94 82 86 98 92 85 83 97 93 82 81 n = 8 M = 93.75 n = 8 M = 83.50 SS = 82 SS = 127.5 %3D %! Is this a one-tailed or two-tailed hypothesis? What is your degrees of freedom? What is your effect size? Express to the nearest hundredths.