10.In a study conducted to determine the role that sleep disorders play in academic performance, researcher JaneGaultney conducted a survey of 503 randomly selected students with a known sleep disorder (such as narcolepsy,insomnia, or restless leg syndrome) and found their mean GPA was 2.65 with a standard deviation of 0.87. Similarly,she randomly selected 1342 students with no known sleep disorders, and found their mean GPA to be 2.82 with astandard deviation of 0.83. You may assume that both population GPA's are normally distributed. Use a significancelevel of 0.05 to test whether the two groups have different mean GPA's.As a follow-up, construct and interpret a 95% confidence interval for the true difference in the mean GPA's of the twogroups.What would be the type I and type II errors for this problem?

PART A SOLUTION: Step 1: State the Null and Alternative HypothesesNull hypothesis:In symbols:Ho: µ1…

Answered: 10. In a study conducted to determine…

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.
A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." A random sample of n1 = 44 adults were asked about fishing as an outdoor activity. The mean response was x1 = 4.9. Another random sample of n2 = 49 adults were asked about camping as an outdoor activity. For this group, the mean response was x2 = 4.1. From previous studies, it is known that ?1 = 1.5 and ?2 = 2.0. Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a 5% level of significance. Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement.
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 703.8. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 36 high-income individuals and found the sample mean credit score to be 719.8 with a standard deviation of 81.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the α=0.05 level of significance.
2. Interpret this data.
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 700.1. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 45 high-income individuals and found the sample mean credit score to be 713.4 with a standard deviation of 84.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) e%3D Make a conclusion regarding the hypothesis. the null hypothesis. There…
A study from 2014 found that adults in the US spend an average of 11 hours per day watching, reading, listening to or simply interacting with media. Researchers at UCLA suspect students spend less time engaging media. They grouped UCLA students by class section, randomly selected 20 class sections, and administered a survey to each student in those randomly selected class sections. The following data represents the mean time (in hours) per day spent engaging media for each of the randomly selected classes: 9.5 11.5 6.0 7.5 8 9.5 10.5 11 12 8.2 5 9.5 8.2 9.3 10.8 11 13 8.6 9.5 10.5 What would be the 99% Confidence Interval for mean time (in hours) spent per day engaging media for UCLA students. Do this "by hand", using appripriate symbols/formulas/tables.
According to the U.S. Department of Transportation’s Air Travel Consumer Report, the nation’s 12 largest airlines recorded an on-time arrival percentage of 77.4% in 2001. Of interest is to estimate the mean delay time for the 22.6% of all flights that did not arrive on time during 2013. A simple random sample of 28 late arriving flights was selected, and the mean delay time of this sample of 28 flights was 14.2 minutes, with a sample standard deviation of s= 6.4 minutes. Use this information to calculate and interpret a 98% confidence interval for the mean delay time for all flights that did not arrive on time during 2013.
Dental health issues remain a barrier to attendance, as evidenced by a 2010 Los Angeles School Attendance Review Board (SARB), which lists dental health, along with homelessness, in the top barriers to attendance. The Dental Hygiene program at West Los Angeles College collects data on its clinic patients to help instructors and students better understand their patient population. From July 2017 to December 2019 the program treated 3421 patients. Students were asked to analyze a random sample of 127 patients from that time period and found that 36.4% had severe periodontal disease (infections and inflammation of the gums and bone that surround and support the teeth). Identify the population: Identify the sample: Identify the parameter of interest: Identify the statistic:
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 704.9. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 33 high-income individuals and found the sample mean credit score to be 717.9 with a standard deviation of 83.9. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the g = 0.05 level of significance. State the null and alternative hypotheses. Ho: µ 704.9 H1: µ > 704.9 (Type integers or decimals. Do not round.) Identify the t-statistic, to = (Round to two decimal places as needed.)
A credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. Values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. According to a survey, the mean credit score is 708.5. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 43 high-income individuals and found the sample mean credit score to be 723.3 with a standard deviation of 84.6. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the a = 0.05 level of significance. State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round.) Identify the t-statistic. to = (Round to two decimal places as needed.) Identify the P-value. P-value = (Round to three decimal places as needed.) Make a conclusion regarding the hypothesis. the null hypothesis. There…
A study from 2014 found that adults in the US spend an average of 11 hours per day watching, reading, listening to or simply interacting with media. Researchers at UCLA suspect students spend less time engaging media. They grouped UCLA students by class section, randomly selected 20 class sections, and administered a survey to each student in those randomly selected class sections. The following data represents the mean time (in hours) per day spent engaging media for each of the randomly selected classes: 9.5 11.5 6.0 7.5 8 9.5 10.5 11 12 8.2 5 9.5 8.2 9.3 10.8 11 13 8.6 9.5 10.5 What would be the calculator steps to find standard deviaton, rounded to 3 decimal places?
A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." A random sample of n1 = 41 adults were asked about fishing as an outdoor activity. The mean response was x1 = 4.9. Another random sample of n2 = 50 adults were asked about camping as an outdoor activity. For this group, the mean response was x2 = 5.7. From previous studies, it is known that ?1 = 1.6 and ?2 = 1.5. Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a 5% level of significance.Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement. (a) What is the level of significance? .05 State the null and alternate…

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