statistics homework chapter 9 mcclave

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October 18, 2010 Maria Esther Vega Statistics 9.38 The Journal of Deaf Studies and Deaf Education. (Fall 2006) published an article on vocalized laughter among deaf users of American Sign Language (ASL). In videotaped ASL conversations among deaf participants 28 laughed at least one. The researchers wanted to know if they laughed more as speakers (while signing) or as audience members (while listening.) For each of the 28 deaf participants the numbers of laugh episodes as an audience member were determined. One goal of the research was to compare the mean numbers of laugh episodes of speakers and audience members. A) Explain why the data should be analyzed as a paired difference experiment. Data should be analyzed as a paired difference experiment because there are two different experiments done in one population, in order to compare both of them. B) Identify the study’s target parameter. The target parameter are the means of the episodes of laughter. c) The study yielded a sample mean of 3.4 laughter episodes for speakers and a sample mean of 1.3 laughter episodes for audience. Is this sufficient evidence to conclude that the population means are different? Explain There is not enough evidence, the mean have to be tested to support the conclusion. d) A paired difference t-test resulted in t= 3.14 and p- value < .01. Interpret the results in the words of problem. p-value < .01 *At 90%, we reject the Ho

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Page 1: Statistics Homework Chapter 9 MCCLAVE

October 18, 2010

Maria Esther Vega

Statistics

9.38 The Journal of Deaf Studies and Deaf Education. (Fall 2006) published an article on vocalized laughter among deaf users of American Sign Language (ASL). In videotaped ASL conversations among deaf participants 28 laughed at least one. The researchers wanted to know if they laughed more as speakers (while signing) or as audience members (while listening.) For each of the 28 deaf participants the numbers of laugh episodes as an audience member were determined. One goal of the research was to compare the mean numbers of laugh episodes of speakers and audience members.

A) Explain why the data should be analyzed as a paired difference experiment.

Data should be analyzed as a paired difference experiment because there are two different experiments done in one population, in order to compare both of them.

B) Identify the study’s target parameter.

The target parameter are the means of the episodes of laughter.

c) The study yielded a sample mean of 3.4 laughter episodes for speakers and a sample mean of 1.3 laughter episodes for audience. Is this sufficient evidence to conclude that the population means are different? Explain

There is not enough evidence, the mean have to be tested to support the conclusion.

d) A paired difference t-test resulted in t= 3.14 and p-value < .01. Interpret the results in the words of problem.

p-value < .01

*At 90%, we reject the Ho

* At 95%, we reject the Ho

* At 99%, we reject the Ho

9.46 Japanese researchers have developed a compression-depression method of testing electronic circuits based on Huffman coding. (IEICE) The new method is designed to reduce the time required for input decompression and output compression- called the compression ratio. Experimental results were obtained by testing a sample of 11 benchmark circuits (all of different sizes) from a SUN blade 1000 workstation. Each circuit was tested with the standard compression- depression method and new Huffman- based coding method and the compression ratio recorded. The data are given in the next table. Compare

Page 2: Statistics Homework Chapter 9 MCCLAVE

two methods with a 95 percent interval. Which method has the smaller mean compression ratio?

With a 95% confidence we determine the Huffman Coding method has a smaller compression mean ratio.

9.59

Bullying behavior study. School bullying is a form of aggressive behavior that occurs when a students is exposed repeatedly to negative actions from another student. In order to study the effectiveness of an anti-bulling policy at Dutch Elementary schools, a survey of over 2000 elementary school children was conducted. Each student was asked if he or she ever bullied another student. In a sample of 1358 boys, 746 claimed that they had never bullied another student. . In a sample of 1379 girls, 967 claimed that they had never bullied another student.

a) Estimate the true proportion of Dutch boys who have never bullied another student.

746

1358= .544 = ΡHat

b) a) Estimate the true proportion of Dutch girls who have never bullied another student.

967

1379= .701

c) Estimate the difference in the proportions with a 90% confidence interval. (-.1877, .-1161)

d) Make a statement about how likely the interval you used in part e contains the true difference in proportions. 90% Interval

e) Which group is more likely to bully another student, Dutch boys of Dutch girls.

Page 3: Statistics Homework Chapter 9 MCCLAVE

Dutch boys, they have a smaller proportion of kids who have not bullied another student.

9.68 Refer to the investigation of collusive bidding in the northern Kentucky school milk market, presented in Exercise 9.26. Market allocation is a common form of collusion behavior in a bid rigging conspiracies. Under collusion, the same dairy usually controls the same school districts year after year. Past experience with milk bids in a competitive environment reveals that a typical incumbency rate is .7. That is 70 % of the school districts are expected to purchase their milk from the dairy that won the previous year. Incumbency rates of .9 or higher are strong indicators of collusive bidding. Over the years the years when bid conclusion was alleged to have occurred in northern Kentucky there where 51 potential vendor transitions in the tri-country market and 134 potential vendor transitions in the surrounding market. These values represent the sample sizes n1= 134 n2= 51 for calculating incumbency rates.

a) Estimate the incumbency rate for the tri-country and surrounding milk market. b) Give a practical interpretation of the results. Do they show further support for the bid conclusion theory.

There are not sufficient funds in order to get a 90% confidence interval we need 521 samples.