Exam 175.271.362.473.363.875.881.961.463.994.475.690.185.187.060.271.245.656.042.018.842.488.089.377.875.564.557.979.689.283.052.761.861.761.881.275.081.777.269.982.083.787.8Exam 269.767.783.561.276.758.749.859.588.375.086.280.392.758.526.349.259.332.761.576.028.957.252.361.183.082.756.749.128.085.775.084.172.843.076.750.576.773.0Question:Using the scores of the exams:(a) Draw a scatterplot of Exam 1 v Exam 2(b) Find the equation of the regression line(c) Use the equation to find the missing valuesfrom the original dataset

Given information: Exam 1 Exam 2 75.2 69.7 71.3 67.7 62.4 83.5 73.3 63.8 61.2 75.8…

Answered: Exam 1 75.2 71.3 62.4 73.3 63.8 75.8…

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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Hours Spent in Child Care 7.5 1.8 0.9 8.0 5.2 6.1 3.3 1.0 4.4 1.6 5.2 11.5 11.6 10.1 9.0 7.3 2.2 1.0 5.2 10.7 7.6 4.1 10.1 6.0 10.1 9.9 10.4 7.0 3.8 9.6 4.1 7.2 4.0 8.5 6.0 7.3 5.4 11.5 9.1 6.5
2.59 2.43 2.96 2.80 2.94 3.07 2.65 2.79 2.61 2.90 3.14 2.78 2.50 2.78 2.52 2.60 2.86 2.88 2.78 3.09 2.82 2.72 2.98 2.68 2.87 2.86 2.78 2.73 2.62 2.77 2.90 3.02 2.73 2.87 2.63 2.98 2.52 2.81 2.91 2.71 3.07 2.89 2.93 2.64 2.94 2.81 2.85 2.89 2.79 2.86 A school administrator believes that the mean class GPAs for a given course are higher than the preferred mean of 2.75 at a significance level of 0.05, and checks a randomly chosen sample of 50 classes. Create a histogram, and calculate x¯, the t-statistic, and the p-value. From the histogram, can normality be assumed? YES OR NO X= t= p= since the p-value ( GREATER THAN OR LESS THAN) the significance level 0.05, the null hypotheses is( fails to be rejected or is rejected) (sufficient or insufficient evidence) exist that the mean GPA is greater than 2.75.
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In relation to the Sustainable Development Goals of the United Nations, one of the critical measures being monitored by the World Health Organization is the annual mean concentrations of fine particulate matter of less than 2.5 microns of diameter (or PM2.5, in ug/m3 units] in urban areas. This is the usual indicator of air pollution. As of 2016, the World Health Organization reported that urban areas in the US have an average of 7.55 ug/m3. You believe that this might have declined in the country due to suspension of economic activities (like transportation operations) during the COVID-19 pandemic. Let us assume the variable follows a normal distribution with standard deviation of 0.11 ug/m3. From the 75 US urban areas that are randomly sampled, the mean is found to be 7.51 ug/m3 (for this same PM2.5 concentration). Conduct a hypothesis test at 5% level of significance. (Source: World Health Organization. Concentrations of fine particulate matter (PM2.5). Retrieved from…
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The Condé Nast Traveler Gold List provides ratings for the top 20 small cruise ships. The data shown below are the scores each ship received based upon the results from Condé Nast Traveler's Annual Readers' Choice Survey. Each score represents the percentage of respondents who rated a ship as excellent or very good on several criteria, including Shore Excursions and Food/Dining. An overall score was also reported and used to rank the ships. The highest ranked ship, the Seabourn Odyssey, has an overall score of 94.4, the highest component of which is 97.8 for Food/Dining. Shore Ship Overall Food/Dining Excursions Seabourn Odyssey 94.4 90.9 97.8 Seabourn Pride 93.0 84.2 96.7 National Geographic Endeavor 92.9 100.0 88.5 Seabourn Sojourn 91.3 94.8 97.1 Paul Gauguin 90.5 87.9 91.2 Seabourn Legend 90.3 82.1 98.8 Seabourn Spirit 90.2 86.3 92.0 Silver Explorer 89.9 92.6 88.9 Silver Spirit 89.4 85.9 90.8 Seven Seas Navigator 89.2 83.3 90.5 Silver Whisperer 89.2 82.0 88.6 National Geographic…
Contact us Search Guides Software Age 1000 9. 1. 991 6. 2. 985 3. 3. 982 3. 4 979 3 976 4. 6. 972 マ 968 4. 964 4. 9 959 Find (i) I; a.9590 b.9760 c. 9640 d. 9850 e. 9790 (ii) d7 a. 30 b.40 C. 60 d. 90 e. 50 8.
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Use the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n= 1048 and x = 568 who said "yes." Use a 95% confidence level. Click the icon to view a table of z scores. Standard Normal (z) Distribution a) Find the best point estimate of the population proportion p. .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 (Round to three decimal places as needed.) 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.0 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.1 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.2 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.3 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.4 0.5 .6915 .6950 .6985 7019 .7054 .7088 .7123 .7157 .7190 .7224 0.5 0.6 .7257 .7291 7324 7357 .7389 .7422 .7454 .7486 .7517 .7549 0.6 0.7 .7580 .7611 .7642 7673 7704 .7734 .7764 .7794…
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