The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates thereare no outliers. Complete parts (a) through (c) below.68.0979.9970.42 83.1179.3385.94101.0297.39 OE Click the icon to view the table of critical t-values.(a) Determinepoint estimate for the population mean travel tax.A point estimate for the population mean travel tax is $ 83.16(Round to two decimal places as needed.)(b) Construct and interpret a 95% confidence interval for the mean tax paid for a three-day business trip.Select the correct choice below and fill in the answer boxes to complete your choice.(Round to two decimal places as needed.)O A. One can be % confident that the mean travel tax for all cities is between $ and $.O B. There is a % probability that the mean travel tax for all cities is between $ and $.O C. One can be % confident that the all cities have a travel tax between $ and $O D. The travel tax is between $ and $ for % of all cities.Clear allCheck answerView an exampleGet more help -Help me solve thisSearch or type URL&%24%23723TYUQKFGS'

Answered: Image /qna-images/answer/eb828dc2-a4ad-4ebb-8f20-667e111aaeb2.jpg

Answered: The accompanying data represent the…

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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Secretary of State Roxana Tillerson is worried about malpractice at the 4 processing centers for Green Cards in the country. Green Card applications are given a grade g on a scale 0-100. They are assigned randomly to different centers and thus Tillerson expects all centers to have similar mean scores. She collects the data in the table. #applications 9360 67.27 15.29 1 9870 66.74 14.64 7020 66.94 15.14 3 5040 67.02 14.84 4 rows x 3 columns At this point she just wants to decide whether or not to initiate a deeper investigation into malpractice at all sites. Does she have sufficient evidence for this, at 6% significance? SSTT number (rtol=0.01, atol=0.0001) SSE number (rtol=0.01, atol=0.0001) F. number (rtol=0.01, atol=0.0001) P-val number (rtol=0.01, atol=0.0001) O (a) No, she cannot make a conclusion O (b) Yes, she has enough evidence Select the key assumptions in the test you used. O (a) The populations are independent and therefore their covariances are zero. O (b) All sample sizes…
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Under each Histogram, write a comment about the shape of the distribution (for example, Symmetric, Skewed (left/right), Uniform or Irregular) Students Frequency Frequency of Students 9 1 0 9 8 1 0 2.0 2.40 1 2 Histogram of Students College GPA 9 2.4 6 4 2.56 Histogram of Students Predicted GPA 9 4 2.8 College GPA 3 2 2.72 Predicted GPA 3.2 3 2.88 3 3 2 3.6 3.04
Find the Z-scores that separate the middle 44% of the distribution from the area in the tails of the standard normal distribution. book 4 Player.aspx?cultureld=&theme=math&style=highered&disableStandbyIndicator=true&lassignmentHandles The Z-scores are (Use a comma to separate answers as needed. Round to two decimal places as needed.) 4 $ Click the icon to view a table of areas under the normal curve. R F F5 通道 % JUDE 5 6 T OL F6 G 1 19 F7 Y F8 H & 7 00 U J DAR F9 * 00 8 O DELL F10 K 9 Locale=true# F11 O ) L 2 F12 P PrtScr 73°F Sunny [ Insert Clear all 0 Delet Backs
The mean age of two people in a room is 20 years. A 50 year old person enters the room. What is the mean age of the 3 people in the room now 20 30 33.3 35 none of the above
omework Assessment A Dashboard + ent/showtest.php The average final exam score for the statistics course is 76%. A professor wants to see if the average final exam score for students who are given colored pens on the first day of class is different. The final exam scores for the 12 randomly selected students who were given the colored pens are shown below. Assume that the distribution of the population is normal. 70, 60, 54, 59, 73, 87, 52, 86, 56, 75, 52, 73 What can be concluded at the the a=0.10 level of significance level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ?v ?v H1: ? ?v c. The test statistic ?v= (please show your answer to 3 decimal places.) d. The p-value : (Please show your answer to 4 decimal places.) e. The p-value is ? a f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest the population mean is not significantly…
The top 16 winning scores for NFL games are listed below. Find the mean score. Video Tutorial: How to find the mean on a TI-84 72 54 52 54 70 51 56 58 58 52 51 63 51 48 52 61 A. 54 B. 57.1 C. 63.2 D. 56.4
Statistics students believe that the mean score on a first statistics test is 65. The instructor thinks that the mean score is higher. She samples 10 statistics students and obtains the scores: Grades 73.5 63.9 64.3 85.5 62.7 69 88 63.9 65 64.3 Test grades are believed to be normally distributed.Use a significance level of 5%.
The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.42 78.57 68.76 84.62 79.17 87.73 101.93 98.62 D Click the icon to view the table of critical t-values. (a) Determine a point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $ (Round to two decimal places as needed.) Clear all Check answer Tutoring Help me solve this Get more help- MacBook Pro G Search or type URL esc & @ 2# $ del 2 3 4 6. 7 P Q W E T Y tab 11 J K .... >
The IQR represents the spread of the middle 50% of the data. Do you think the IQR in this problem is large or small? Here is the data. 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 7, 13, 14 Median is 3. Min # is 0 Max # is 14 Q1 is 2 Q2 is 3 Q3 is 4 IQR=Q3 (4)-Q1(2) IQR=2
Mean = 64.57 median = 64.50 mode = 64.09 comment on the shape of the distribution. Give reasons for your answer.

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