R5.21. Teach for America Several programs attempt to addressthe shortage of qualified teachers by placing uncertifiedinstructors in schools with acute needs often in inner cities.A study compared students taught by certified teachers toothers taught by uncertified teachers in the same schools.Reading scores of the students of certified teachers averaged35.62 points with standard deviation 9.31. The scores of stu-dents instructed by uncertified teachers had mean 32.48 pointswith standard deviation 9.43 points on the same test. Therewere 44 students in each group. The appropriate 1-procedurehas 86 degrees of freedom. Is there evidence of lower scoreswith uncertified teachers? Discuss. (The Effectiveness of"Teach for America" and Other Under-certified Teacherson Student Academic Achievement: A Case of HarmfulPublic Policy. Education Policy Analysis Archives)

Given: x¯1=35.62s1=9.31x¯2=32.48s1=9.43n1=n2=44 Degrees of freedom(df): df=86 X follows two sample…

Answered: R5.21. Teach for America Several…

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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In a study examining the relationship between socioeconomic status (SES) and academic achievement among high school students, the researchers collected data on 500 students from varying SES backgrounds. The SES was categorized into low, medium, and high. The academic achievement was measured using GPA scores. The following descriptive statistics were obtained: Mean GPA for low SES students: 2.8 Standard deviation of GPA for low SES students: 0.5 Mean GPA for medium SES students: 3.5 Standard deviation of GPA for medium SES students: 0.3 Mean GPA for high SES students: 4.0 Standard deviation of GPA for high SES students: 0.2 a) Calculate the coefficient of variation for each SES group and interpret the results.
9. You are conducting an experiment on the effectiveness of a reading intervention with young children and want to report the descriptive results to your colleagues. Some of the children received the inter- vention, and some did not. What do the averages and standard deviations tell you about the results? For this exercise, use the data in Chapter 3 Data Set 2. No Intervention 18 16 17 15 16 14 22 21 7 24 23 19 12 20 23 Intervention 17 15 21 24 12 16 14 25 19 24 21 25 9 9 13
Fuel efficiency of Prius: Fueleconomy.gov, the official US government source for fuel economy information, allows users to share gas mileage information on their vehicles. The histogram below shows the distribution of gas mileage in miles per gallon (MPG) from 14 users who drive a 2012 Toyota Prius. The sample mean is 53.3 MPG and the standard deviation is 5.2 MPG. Note that these data are user estimates and since the source data cannot be verified, the accuracy of these estimates are not guaranteed. (a) The EPA claims that a 2012 Prius gets 50 MPG (city and highway mileage combined). Do these data provide strong evidence against this estimate for drivers who participate on fueleconomy.gov? The test statistic is: __________________(please round to two decimal places)The p-value:___________________________ for this hypothesis test is: (please round to four decimal places)(c) Please comment on the primary thing that is wrong about the following statement: There is sufficient evidence…
5. The heights of a pediatricians three-year-old females are approximately normally distributed, with mean 38.72 inches and standard deviation 3.17 inches. Find the height of a three-year-old female at the 42nd percentile.
1. The EPA fuel economy estimates for automobile models tested recently predicted a mean of 24.8 mpg and a standard deviation of 6.2 mpg for highway driving. a. About what percent of cars get less than 31 mpg? b. What percent of cars get more than 20 mpg? c. What percent of cars get between 22 and 30 mpg? d. What gas mileage represents the 75th percentile? e. What gas mileages separate the bottom 3% and top 3% of cars? X
bcd.
The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students' study habits and attitudes toward school. The survey yields several scores, one of which measures student attitudes toward studying. The mean student attitude score for college students is about 30, and standard deviation is about 12. A researcher in the Philippines is concerned about the declining performance of college graduates on professional licensure and board exams. She suspects that poor attitudes of students are partly responsible for the decline and that the mean for college seniors who plan to take professional licensure or board exams is less than 30. She gives the SSHA to an SRS of 144 college seniors in the Philippines who plan to take professional licensure or board exams. Suppose we know that the student attitude scores in the population of such students are Normally distributed with standard deviation σ = 12. (a) We seek evidence against the claim that μ = 30. What is the…
Answer.
Beer Drinking. The mean annual consumption of beer per person in the US is 22.0 gallons . A random sample of 300 Washington D.C. residents yielded a mean annual beer consumption of 27.8 gallons. At the 10% significance level, do the data provide sufficient evidence to conclude that the mean annual consumption of beer per person for the nation’s capital differs from the national mean? Assume that the standard deviation of annual beer consumption for Washington D.C. residents is 55 gallons.
MC Qu. 10-64 Sales at a fast-food restaurant average... Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to Increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is higher than $6,000 per day. Reject the null hypothesis and conclude the mean is lower than $6,000 per day. Reject the null hypothesis and conclude that mean equal to $6,000 per day.

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