Use the following sample to estimate a population mean u.32.129.950.925.636.921.428.326.128.834.646.826.920.329.736.228.444.522.348.131.326.338.425.513.517.124.618.234.714.531.333.624.124.926.418.127.623.741.28.323.839Download CSVFind the 99.5% confidence interval about the population mean. Enter your answer as atri-linear inequality accurate to two decimal place.

Answered: Use the following sample to estimate a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Use the following sample to estimate a population mean μμ. 46.6 32.3 36.1 39.6 34.8 47.9 44.7 20.5 17.8 53.1 63.7 40 37.1 54.9 53.5 40.9 34.6 63.3 12.8 42.4 39.9 57 1.5 43.2 68.2 33.8 16.2 25.8 45.8 32 27.2 45.2 8.3 52.9 29.8 22.4 40.5 23.2 58.3 42.1 47.9 38 Download CSV Find the 99% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. < μμ <
6. 6. 4 9 For the sample data in the table above, by how much does the data value 4 deviate from the mean? Round to one decimal if needed and enter a positive value. Submit Question MacBook Pro G Search or type URL 6O 7 6
Here is a sample data set that appears to be nearly normal (as suggested by the histogram). 85.4 63.6 85.7 85.6 92.4 89.8 84.2 82.2 88 76.5 87.6 79 61.4 88.6 89.4 95 96.4 76.3 92.4 79.8 82.2 86.5 76.7 90.7 85.4 94 75.8 79.8 92.4 74.3 65 81.4 75.5 76.7 76.5 84.2 78.6 82.9 84.7 74.9 84 91.2 84.4 68.2 76 75.8 74.6 77.3 79.8 69.2 82.9 74.6 71.7 76.7 80 82.2 91.8 74.9 72.7 74.8 84 80.6 67.6 71.7 79.6 81.2 78.2 83.7 82.2 71.4 What is the mean of this data set? (Report answer using the rules suggested in class: 1 more d.p. than the data.)M=What is the standard deviation of this data set? (Report answer using the rules suggested in class: 2 more d.p. than the data.)SD=
Here is a sample data set (n = 48) that is nearly normal. 51 52 44.4 46.9 48.2 48.8 52 52.4 52.4| 53.4 51 53.5 54.3 55.1 57.3 58.2 58.4 58.6 58.6 58.8 59.2 60.5 60.8 61.4 61.8 62.7 63.3 63.5 64.2 65.7 65.7 66.2 66.9 67.6 69.8 70.3 70.7 72.7 73.2 67 68.3 74 How many outliers are in this data set? ans = 52 57.3 59.8 63.5 68 77.8
Use the following sample to estimate a population mean µ. 11.7 6.6 22.6 41.4 26.9 -0.7 30.2 -7.9 -1.1 20.8 13.3 46.5 -13.4 30.2 6.2 17.3 19 13.1 9.7 17.7 1.7 26.1 2.1 57.8 40.3 46.4 9.4 7.5 11.3 -5.8 1.3 -0.5 12.1 19.2 15.6 25.4 24.4 50.7 0.4 6.8 24.2 24.8 24.5 Download CSV Find the 99% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place. < µ< N327 -43 4
4. Given the following 25 sample observations: 5.3, 6.1, 6.7, 6.8, 6.9, 7.2, 7.6, 7.9, 8.1, 8.9, 9.0, 9.2, 9.4, 9.7, 10.1, 10.4, 10.6, 10.8, 11.3, 11.4, 12.0, 12.1, 12.3, 12.5, 13.2 And let Y₁, Y2,, Y be the order statistics for this sample A) The interval (Y9, Y16) could serve as distribution-free estimate of the median, m, of the population. Find the confidence coefficient of this interval. (6 points) B) The interval (Y3, Y10) be a confidence interval for π 25. Determine this confidence interval and, Using Table II in Appendix B of the textbook. Determine the confidence coefficient of this interval.
Here is a sample data set. DO THIS BY HAND, WITHOUT USING TECHNOLOGY. 334.9 337.5 340.2 352.8 356.2 356.3 360.3 361.4 363.9 366.6 367.9 369.7 371.4 371.4 371.4 371.5 371.8 372.3 373.4 373.5 375.5 375.9 378.9 379 380.3 380.3 381.3 381.3 384.9 387.4 388.7 389 392.1 393.3 393.7 394.6 394.6 396.6 397.4 397.5 397.5 397.5 400.1 406.1 420.2 429.1 429.5 430 432 441.2 445 445.9 447.3 449.7 Find the first quartile for this data set.Q1 = Find the third quartile for this data set.Q3 = Find the interquartile range for this data set. IQR = Q3 - Q1IQR =
Here is a sample data set. 400 404.8 409.1 409.5 413.1 413.8 418.5 424.7 428.9 432.3 436.1 437.5 440 440 440 441.1 442.4 444.3 445.6 446.1 448 448.8 449.7 450.9 451 453.2 462.3 462.3 462.7 463.6 464.2 465.5 466.7 466.7 467.2 467.7 470.3 472.3 472.6 473.3 473.3 473.3 478.7 479.1 479.3 480.8 485.1 486 490.4 493.1 495.2 496 501.7 506.6 5 10 15 20 25 length (cm) 400 450 500 550 Frequency [Graphs generated by this script: setBorder(54,40,20,15); initPicture(355,550,0,29);axes(1013.2,5,1,null,5); fill="blue"; stroke="black"; textabs([165,0],"length (cm)","above");line([400,-0.58],[400,0.58]); text([400,0],"400","below");line([450,-0.58],[450,0.58]); text([450,0],"450","below");line([500,-0.58],[500,0.58]); text([500,0],"500","below");line([550,-0.58],[550,0.58]); text([550,0],"550","below");textabs([0,115],"Frequency","right",90);rect([400,0],[450,23]);rect([450,0],[500,29]);rect([500,0],[550,2]);] Find the first quartile for this data set. Q1 = 440 Correct Find the third quartile for…
Here is a sample data set. 278.2 282.5 311.5 311.6 320.4 322.9 352.3 367.4 375.7 390.8 395.8 411.7 412.9 412.9 412.9 418.6 419 420.6 423.1 425.2 426.9 429.1 429.6 433.5 434.3 434.8 435.4 435.4 436.2 444 451.7 453 456.5 464.1 473.2 476.1 478.2 489.2 497 502.7 502.7 502.7 513.5 517.9 522.2 566.6 568.3 576.1 577.1 580.8 604.5 618.9 621.6 628.4 24681012141618length (cm)250300350400450500550600650Frequency[Graphs generated by this script: setBorder(54,40,20,15); initPicture(205,650,0,19);axes(1256.8,2,1,null,2); fill="blue"; textabs([165,0],"length (cm)","above");line([250,-0.38],[250,0.38]); text([250,0],"250","below");line([300,-0.38],[300,0.38]); text([300,0],"300","below");line([350,-0.38],[350,0.38]); text([350,0],"350","below");line([400,-0.38],[400,0.38]); text([400,0],"400","below");line([450,-0.38],[450,0.38]); text([450,0],"450","below");line([500,-0.38],[500,0.38]); text([500,0],"500","below");line([550,-0.38],[550,0.38]);…
Use the following sample to estimate a population mean . 70.6 77.8 65.1 72.1 57.7 53.6 40.6 53.5 38.9 68.1 99 68.8 64.8 59.3 89.2 52.4 100.9 97.9 47.7 60.3 52.4 63.5 65.2 119.4 62.5 103.5 90.1 88.1 49.8 61.2 81.9 20.9 72 59.1 83.2 73.8 62.8 119.9 55.1 88.5 103.2 86.6 Download CSV 55.7 96.3 115.9 Find the 80% confidence interval about the population mean. Enter your answer as a tri-linear inequality accurate to two decimal place.
The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a "flipped" classroom. Complete parts (a) through (c) below. Traditional 69.9 69.6 78.7 67,6 84.7 78.5 56.3 80.2 81.9 71.0 64.4 70.2 59.0 Flipped 77.5 72.5 62.7 72.6 76.9 90.4 79.6 77.4 81.9 70.5 91.3 77.8 75.7 (a) Which course has more dispersion in exam scores using the range as the measure of dispersion? The traditional course has a range of 28.4 , while the "flipped" course has a range of 28.6. The flipped course has more dispersion. (Type integers or decimals. Do not round.) (b) Which course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion? The traditional course has a standard deviation of, while the "flipped" course has a standard deviation of The V course has more dispersion. (Round to three decimal places as needed.)
A magazine collects data each year on the price of a hamburger in a certain fast food restaurant in various countries around the world. The price of this hamburger for a sample of restaurants in Europe in January resulted in the following hamburger prices (after conversion to U.S. dollars). 5.18 4.96 4.07 4.68 5.24 4.67 4.15 4.98 5.16 5.56 5.33 4.60 The mean price of this hamburger in the U.S. in January was $4.61. For purposes of this exercise, assume it is reasonable to regard the sample as representative of these European restaurants. Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price? Test the relevant hypotheses using ? = 0.05. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to four decimal places.) t = P-value =

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