Can you please help me with one multiple choice.Use excel please if you have to. i reallyappreciated.The Student Government Association at Middle CarolinaUniversity wanted to demonstrate the relationship betweenthe number of beers a student drinks and his or her bloodalcohol content (BAC). A random sample of 18 studentsparticipated in a study in which each participating studentwas randomly assigned a number of 12-ounce cans of beerto drink. Thirty minutes after they consumed their assignednumber of beers, a member of the local sheriff's officemeasured their blood alcohol content. The sampleinformation is reported below.Click to download the data set: MiddleCarolina.xlsxWhat is the expected BAC for someone who drinks a 12-pack of beer?0.160.20Cannot be determined0.10 1.StudentBeersBAC6.0.1070.09370.09440.105.0.1076.30.078.70.109.8.6.0.12109.6.0.09111030.07121130.05131270.08141310.04151440.07161520.06171670.1218170.0519180.02202184LO

Explanation: Enter the data in MS Excel Select the columns Beer and BAC Go to Insert tab Click on…

Answered: The Student Government Association at…

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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Determine the sample in this research scenario: Dr. Eglovitch wants to know whether type of exercise, running or yoga, improves post-partum depression. One hundred women are recruited from obstetricians’ offices in the Richmond area. Group of answer choices A.100 postpartum women whose depression was measured B.Type of exercise C. Postpartum depression D. All postpartum women
the number of observations and measurements taken is the sample size True or False
Consider a sample with data values of 27, 25, 20, 15, 30, 34, 28, and 25. Compute the 19th, 24th, 65th, and 75th percentiles. If needed, round your answers to two decimal digits. Percentile Value 19% 24% 65% 75%
1. Explain the differences between a sample and a population.
During the first 13 weeks of the television season, the Saturday evening 8:00 PM to 9:00 PM audience proportions were recorded as ABC 29%, CBS 26%, NBC 24%, and Independents 21%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 93 homes, CBS 68 homes, NBC 87 homes, and independents 52 homes. Test with a .05 to determine whether the viewing audience proportions changed. Round your answers to two decimal places. Test statistic 6.76 ✔ p-value is between between 05 and.101 Conclusion: There is no significant change in the viewing audience proportions.
Part b should be a sample size of 29%, not 25%
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gave the following data for home run percentages.
Please help me in answering the following practice question:Given a random sample of the amount of protein (in gram) in one bag of snack food: 0.95, 0.85, 0.92, 0.95, 0.93, 0.86, 1, 0.92, 0.85, 0.81, 0.78, 0.93, 0.93, 1.05, 0.93, 1.06, 1.06, 0.96, 0.81, 0.96 Here, it is assumed that the amount of protein in the snack is normal. How would I calculate a 98% two-sided confidence interval for the average amount of protein in one bag?Thank you for your help in advance!
Betty, an employee of Shining Sun Daycare Centerread an article in Healthy Child Magazine sayingthat the average 3-year-old child is 37 in. tall.Betty works with 10 3-year-olds at Shining Sun,so later that week, she measured the height ofeach child who had just turned or was about toturn 3 years old. Here are the descriptive statisticsfor the sample: M = 35, s = 3.1. State the nondirectional hypothesis.2. Determine the critical t for a = .05.3. Calculate t. Show your calculations.4. Is the height of 3-year-olds in Shining SunDaycare Center significantly different fromthe height given in the magazine?
A college student wants to determine whether the stress level of a student is associated to his/her body mass index (BMI). She randomly selected 80 students and determine their stress level and BMI. The data obtained is summarized in the table below. BMI Stress Level Total Normal Not Normal Low 20 31 51 Hight 15 14 29 Total 35 45 80
PrActicE the legalization of marijuana. Among the Democratic voters surveyed, 158 of the 200 people aged 18-29 and 210 of the 300 people aged 50-64 supported legalizing marijuana. Is this difference significant at a = .05? A random sample of individuals in one town was surveyed about their position on
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