AutoSaveFile19A8 C9d34 a35 bHome⠀HInsert Draw Page LayoutReadyB$105,856$104,905H$2,087$1,999X ✓ fx$484.52$464.10C2627282930313233 19 The point estimate of the mean repair cost is,$2,274$2,179$444.54$425.80132021252305165515401485D36 C37 d3839 20 The standard deviation of the sample data is,240 a241 b4-=137021852950202520052305Accessibility: InvestigateE270 HW5 LWP WEEDType here to searchE2021 Questions 19-22 are related to the following22 Your company self-insures its fleet of cars against collisions. You need to estimate the mean repair costs per accident so23 that you can set aside enough money to cover these costs. The data for repair cost for a random sample of accidents is24 shown below.25215519702050202019901590E270 HW5 FA22 ✓Formulas Data+F196510752095183026502690G242 C243 d244245 21 The margin of error for an interval estimate with a 95% level of confidence is,246 a$129.08O25951950965199022851860O Search (Alt+())ReviewH21 7 nView234519201455217018402135HelpPower Pivot187013052645237019401720K276023202790205516701640 FileK194146 a47 bHome48 с49 d5051 225253 a54 b55Y :ABСDE39 20 The standard deviation of the sample data is,40 ab42C43 dLeadyH$484.52$464.10$444.54$425.80$129.08$134.46Insert$140.06$145.90Draw Page LayoutX ✓1311461621804445 21 The margin of error for an interval estimate with a 95% level of confidence is,FormulasFWhat was the sample standard deviation?$425.81$417.46$409.27$401.25Accessibility: InvestigateType here to search14+Data$1,884.91 $2,122.59GReview ViewTo obtain a margin of error of $65 with a 95% level of confidence, what is the minimum number of repair costs tobe included in the sample? Use standard deviation from the sample data as the planning value.с56 d5758 23 A different interval estimate of the mean repair cost, using the sample size as the original sample, yielded thefollowing 95% confidence interval.59606162 a63 b64C55 d6657 Use the binary data in the worksheet tab LWP to answer Questions 24-28.E270 HW5 LWP WEEDOHHelp3'Power PivotKL O

Answered: Image /qna-images/answer/0c8126ac-6fb5-443c-9f72-2c76007dca76.jpg

9 20 21 Questions 19-22 are related to the following 22 Your company self-insures its fleet of cars against collisions. You need to estimate the mean repair costs per accident so 23 that you can set aside enough money to cover these costs. The data for repair cost for a random sample of accidents is 24 shown below. 25 26 27 28 29 30 31 32 d $104,905 34 a 35 b 36 с d 78 37 33 19 The point estimate of the mean repair cost is, $2,274 $2,179 $2,087 $1,999 242 C 243 d 244 245 21 246 a 1320 2125 Ready 2305 1655 1540 1485 $484.52 $464.10 $444.54 $425.80 238 239 20 The standard deviation of the sample data is, 240 a 241 b 1370 2185 2950 2025 2005 2305 2155 1970 2050 2020 1990 1590 E270 HW5 LWP WEED Accessibility: Investigate Type here to search 1965 1075 2095 1830 2650 2690 2595 1950 965 1990 2285 1860 The margin of error for an interval estimate with a 95% level of confidence is, $129.08 2345 1920 1455 2170 O E 1840 2135 1870 1305 2645 2370 1940 1720 2760 2320 2790 2055 1670 1640 L Ⓡ