The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $24,100. You take a random sample of 136 college students in the state of Vermont and find the mean debt is $25,000 with a standard deviation of $2,400. We want to construct a 90% confidence interval for the mean debt for all Vermont college students.

(a) What is the point estimate for the mean debt of all Vermont college students?

(b) What is the critical value of t (denoted tα/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 3 decimal places. tα/2 =

(c) What is the margin of error (E) for a 90% confidence interval? Round your answer to the nearest whole dollar. E = $

(d) Construct the 90% confidence interval for the mean debt of all Vermont college students. Round your answers to the nearest whole dollar. < μ <

(e) Based on your answer to (d), are you 90% confident that the mean debt of all Vermont college students is greater than the quoted national average of $24,100 and why?

a. No, because $24,100 is above the lower limit of the confidence interval for Vermont students.

b. Yes, because $24,100 is above the lower limit of the confidence interval for Vermont students.

c. Yes, because $24,100 is below the lower limit of the confidence interval for Vermont students.

d. No, because $24,100 is below the lower limit of the confidence interval for Vermont students.

(f) We are never told whether or not the parent population is normally distributed. Why could we use the above method to find the confidence interval?

a. Because the sample size is less than 100.

b. Because the margin of error is positive.

c. Because the sample size is greater than 30.

d. Because the margin of error is less than 30.

1. The provost at the University of Chicago claimed that the entering class this year is larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year’s entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The University’s record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. Round final answers to two decimal places. Solutions only.

(A) The parameter the president is interested in is:
(a) the mean number of entering students to his university this year.
(b) the mean number of entering students to all U.S. universities this year.
(c) the mean SAT score of the entering students to his university this year.
(d) the mean SAT score of the entering students to all U.S. universities this year.

(e) None of the above.

(B) The population the president is interested in is:
(a) all entering students to all universities in the U.S this year.

(b) all entering students to his university this year.
(c) all SAT test centers in the U.S. this year.
(d) the SAT scores of all students entering universities in the U.S. this year.

(e) None of the above.

(F) True, False, or Uncertain: The null hypothesis would be rejected.

(G) True, False, or Uncertain: The null hypothesis would be rejected if a 10% probability of committing a Type I error is allowed.

(I) True, False, or Uncertain: The evidence proves beyond a doubt that the mean SAT score of the entering class this year is lower than previous years.

(J) True, False, or Uncertain: If these data were used to perform a two-tail test, the p-value would be 0.1254.

Student Debt – Vermont: The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to be about $22,500. You take a random sample of 136 college students in the state of Vermont and find the mean debt is $23,500 with a standard deviation of $2,600. We want to construct a 90% confidence interval for the mean debt for all Vermont college students.

(a) What is the point estimate for the mean debt of all Vermont college students?

(b) What is the critical value of t (denoted t_α/2) for a 90% confidence interval? Use the value from the table or, if using software, round to 3 decimal places.
t_α/2 =

(c) What is the margin of error (E) for a 90% confidence interval? Round your answer to the nearest whole dollar.
E = $

(d) Construct the 90% confidence interval for the mean debt of all Vermont college students. Round your answers to the nearest whole dollar.
< μ <

(e) Based on your answer to (d), are you 90% confident that the mean debt of all Vermont college students is greater than the quoted national average of $22,500 and why?

Yes, because $22,500 is below the lower limit of the confidence interval for Vermont students

.No, because $22,500 is above the lower limit of the confidence interval for Vermont students.   

Yes, because $22,500 is above the lower limit of the confidence interval for Vermont students.

No, because $22,500 is below the lower limit of the confidence interval for Vermont students.

(f) We are never told whether or not the parent population is normally distributed. Why could we use the above method to find the confidence interval?

Because the sample size is greater than 30.

Because the sample size is less than 100.    

Because the margin of error is less than 30.

Because the margin of error is positive.

Methods

Participants

Participants (N = 8,997) were undergraduate students at 20 2- and 4-year college and university campuses in Minnesota who completed the 2015 College Student Health Survey.¹⁷ Data were collected online between February 16, 2015 and March 27, 2015. Most (68%) participants were female. Participants identified as White (81.6%), Asian (8.6%), Black or African American (6.1%), American Indian or Alaska Native (2.1%), Native Hawaiian or other Pacific Islander (0.4%), other races (1.9%), and prefer not to answer (2.9%). Percentages did not add to 100% because respondents could indicate more than one race. The median age of the sample was 21 years old (range =18–74). Most (73%) of the sample was enrolled in 4-year colleges and universities. Almost half (47%) were first-generation college students (ie, neither parent had a 4-year college degree). The sample included participants who identified as heterosexual (89%), bisexual (4%), g** or lesbian (2%), or other (4%). The study was approved by the institutional review board (IRB) of the university that sent the survey (Study # 0012S75881). Other schools signed an agreement with that university or sought approval from their own IRB. All study procedures were carried out in accordance with the latest version of the Declaration of Helsinki. Students provided consent on the online survey.

Measures

Factors perceived to affect academic performance

Factors that were perceived to affect college academic performance were assessed by a measure that presented students with a list of 20 factors, including stress.¹⁷Students rated each factor in terms of whether they had experienced it and, if so, whether or not it had affected their academic performance during the past 12 months (1 = I do not have this issue/not applicable; 2 = I have this issue, my academics have not been affected; 3 = I have this issue, my academics have been affected). The similar question on the NCHA survey assessed how academics were affected (eg, scored lower on a test or project) in separate responses, graded for severity of the academic impairment, whereas this survey had a single response option for academics being affected by the factor.

Stress

Stress was measured on a 10-point scale on which students rated their average level of stress over the past 30 days (1 = Not stressed at all to 10 = Very stressed). The construct validity of scores on this measure was supported by the finding that students who scored higher on this measure also reported more days with poor mental health (including stress, depression, and problems with emotions) in the past month (r = .49, p < .001).

Coping self-efficacy

Coping self-efficacy was measured on a 10-point scale on which students rated their ability to manage their stress over the past 30 days (1 = Ineffective to 10 = Very effective). The construct validity of scores on this measure was supported by a negative correlation between coping self-efficacy and stress (r = −.47, p < .001).

Resilience

The Brief Resilience Scale (BRS)¹⁸ is a 6-item measure of resilience, defined as the ability to “bounce back or recover from stress” (p. 194). Students rated each item (eg, “It does not take me a long time to recover from a stressful event”) on a 5-point scale (1 = Strongly Disagree to 5 = Strongly Agree). Higher scores indicate greater resilience. Smith et al reported a test-retest reliability coefficient of .69 in a sample of undergraduate students over a one-month period and Cronbach’s alphas ranging from .84 to .87. In the current sample, α was .88.

Social support

Students rated the supportiveness of their family, friends, college/university faculty, and college/university staff on four separate items. Each item was rated on a 10-point scale (1 = Very unsupportive to 10 = Very supportive). To create an overall measure of social support, scores on these four items were averaged (α = .84).

Academic performance

Participants reported their cumulative grade point average (GPA) to two decimal places (M = 3.35, SD = 0.49). The correlation between self-reported college GPA and data from student records was r = .90 in a meta-analysis.¹⁹

^{__________________________________________________________________________________________________________________________________________________________}

^{Please read above reference and then...}

^{Q. Research Methods}

Discuss the methods the researchers used for: selecting study participants and collecting and analysing data.

What were the strengths and weaknesses of the methods used?

TzeMay was one of the first women engineering students at ABC University. She graduated in 1995 with a first class honours degree and immediately continued her studies with an MSc programme, gaining recognition for her work into environmentally friendly car engines, a largely untapped field in those days. On completion of her Masters degree she was offered a post as a research assistant where she could have developed her Masters research and worked towards her doctorate. However she decided that she needed to gain some commercial experience and joined Wallace-Price, a blue-chip engineering consultancy where, apart from a sponsored year out to study for an MBA in the United States of America, she has remained ever since.

Her tenacity and loyalty to Wallace-Price have paid off and she was made a partner in the firm, primarily responsible for bringing in work to the consultancy. With the promotion came various executive privileges including an annual salary of £80, 000, a chauffeur-driven car, free use of one of the company-owned London flats, a non-contributory pension scheme, various gold credit cards and first-class air travel. TzeMay herself would not describe these as benefits, however, but as necessities to enable her to do her job properly. In order to meet her business target of £2 million of work for Wallace-Price she spent forty weeks overseas, working an average of ninety hours a week.

She cannot remember the last time that she had a weekend when she was not entertaining clients or travelling but was totally free to indulge herself. During her time with Wallace-Price she has earned a reputation both as a formidable but honest negotiator and as an innovative engineer, often finding seemingly impossible solutions to problems. Known for her single-minded dedication to her job, she does not suffer fools gladly. She is frequently approached to work for rival firms with promises of even greater privileges and has been the subject of numerous magazine profiles, some concentrating on her work and reputation as a high flier but the majority focusing on her gender. Her fortieth birthday last year was spent alone in the Emergency Room of a Los Angeles hospital where she had been rushed with a suspected stomach ulcer. Deprived of her portable telephone, fax and computer she had little else to do but to reflect on her life thus far. On her return to health she was working her way through the pile of technical journals, which had accumulated during her absence and there she saw the advertisement for ABC University, an institution that had close links with her company and whose Professor of Engineering she knew well. Ignoring the instructions relating to applications she put through a telephone call to the ABC University.

Question 1 : Making reference to the appropriate theories of motivation, explain TzeMay‟s main motivating factors.

C++

Write a program that prompts the user to enter 50 integers and stores them in an array. The program then determines and outputs which numbers in the array are sum of two other array elements. If an array element is the sum of two other array elements, then for this array element, the program should output all such pairs separated by a ';'.

An example of the program is shown below:

list[0] = 15 is the sum of: 
----------------------
list[1] = 25 is the sum of: list[5], list[14]; list[13], list[47]; list[46], list[47]; 
----------------------
list[2] = 23 is the sum of: list[0], list[33]; 
----------------------
list[3] = 65 is the sum of: list[4], list[32]; list[13], list[20]; list[20], list[46]; list[21], list[34]; list[32], list[38]; 
----------------------
list[4] = 34 is the sum of: list[0], list[14]; list[6], list[48]; 
----------------------
list[5] = 6 is the sum of: 
...

If no two elements can equal the value of an element, the line can remain empty, as shown above for list[0] and list[5].

Instructions

Write a program that prompts the user to enter 50 integers and stores them in an array. The program then determines and outputs which numbers in the array are sum of two other array elements. If an array element is the sum of two other array elements, then for this array element, the program should output all such pairs separated by a ';'.

An example of the program is shown below:

list[0] = 15 is the sum of: 
----------------------
list[1] = 25 is the sum of: list[5], list[14]; list[13], list[47]; list[46], list[47]; 
----------------------
list[2] = 23 is the sum of: list[0], list[33]; 
----------------------
list[3] = 65 is the sum of: list[4], list[32]; list[13], list[20]; list[20], list[46]; list[21], list[34]; list[32], list[38]; 
----------------------
list[4] = 34 is the sum of: list[0], list[14]; list[6], list[48]; 
----------------------
list[5] = 6 is the sum of: 
...

If no two elements can equal the value of an element, the line can remain empty, as shown above for list[0] and list[5].

C++

among 3000 college students 13% are over 21 years old what is the probability that out of a sample of 550 students at least 15% are 21 years old