Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students. The scores areshown below44 50 10 92 76 35 64 36 43 72 54 62 35 74 5072 36 29 39 61 48 63 35 41 22 36 50 46 85 13To find the 10% trimmed mean of a data set order the data, delete the lowest 10% of the entries and the highest 10% of the entries, and find the mean of theremaining entries. Complete parts (a) through (c)CEC(a) Find the 10% trimmed mean for the dataThe 10% trimmed mean is(Round to the nearest tenth as needed)

Answered: Students in an experimental psychology…

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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The midterm and final exam grades for a statistics course are provided in the data set below. Jaymes, a student in the class, scored 86 on both exams. Treat the given data sets as samples. Midterm: 80, 78, 85, 82, 79, 79, 78, 86, 80, 84, 78, 84, 80, 84, 81, 78, 81, 82, 78, 84 Final: 81, 88, 68, 69, 69 81, 82, 86, 76, 71, 72, 68, 77, 86, 68, 83, 84, 71, 81, 87 z-score for midterm: 1.857 z-score for final: 1.181 Based on the z-scores calculated above, which of Jaymes's grades is more unusual, the midterm grade or the final exam grade? Select the correct answer below: A. The absolute value of the z-score for the final exam grade is greater than for the midterm grade, so the final exam grade is more unusual. B. The absolute value of the z-score for the midterm exam grade is less than for the final grade, so the midterm grade is more unusual. C. The absolute value of the z-score for the midterm exam grade is greater than for the final grade, so the midterm grade is more…
A marketing firm is doing research for an internet-based company. It wants to appeal to the age group of people who spend the most money online. The company wants to know if there is a difference in the mean amount of money people spend per morith on Internet purchases depending on their age bracket. The marketing firm looked at two age groups, 18-24 years and 25-30 years, and collected the data shown in the following table. Let Population 1 be the amount of money spent per month on internet purchases by people in the 18-24 age bracket and Population 2 be the amount of money spent per month on internet purchases by people in the 25-30 age bracket. Assume that the population variances are not the same. Internet Spending per Month 18-24 Years Answer Mean Amount Spent Standard Deviation Sample Size Step 1 of 2: Construct a 99 % confidence interval for the true difference between the mean amounts of money per month that people in these two age groups spend on Internet purchases. Round the…
South Dakota prices vary $350,000: $400,000:$400,000:$425,000:$429,000: $440,000: $560,000: $585,000,: $585,000: $785,000 : $1,300,000. We found the following Mean = 569000 SD = 272703.1353 T score = .7113 P Value = .4932 CI: 385795.5409,752204.4591 If you had a margin of error of $13,000 how large of sample would you need with a 90% confidence interval using the given tscore.
The college physical education department offered an advanced first aid course last summer. The scores on the comprehensive final exam were normally distributed, and the z-scores for some of the students are shown below. Robert 1.24 Juan 1.66 Haley −1.84 Joel 0.00 Jan −0.84 Linda 1.68 Which of these students scored above the mean? (Select all that apply.) Which of these students scored on the mean? (Select all that apply.) Which of these students scored below the mean? (Select all that apply.) If the mean score was μ = 140 with standard deviation σ = 15, what was the final exam score for each student?
Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students.The scores are shown below. 43 51 11 91 77 36 64 37 42 72 54 62 36 75 51 D 72 37 29 39 61 48 63 36 41 21 37 51 47 86 14 To find the 10% trimmed mean of a data set, order the data, delete the lowest 10% of the entries and the highest 10% of the entries, and find the mean of the remaining entries. Complete parts (a) through (c). (a) Find the 10% trimmed mean for the data. The 10% trimmed mean is (Round to the nearest tenth as needed.) (b) Compare the four measures of central tendency, including the midrange. Mean = (Round to the nearest tenth as needed.) Median = Mode = (Use a comma to separate answers as needed.) The midrange is found by the following formula. (Maximum data entry) + (Minimum data entry) Midrange = (Round to the nearest tenth as needed.) (c) What is the benefit of using a trimmed mean versus using a mean found using all data…
A sample of 25 freshmen at a major university completed a survey that measured their degree of racial prejudice (the higher the score, the greater the prejudice) Compute the median and mean for these data? 10 43 30 30 45 40 12 40 42 35 45 25 10 33 50 42 32 38 11 47 22 26 37 38 10
A shopper at a local supermarket has spent these amounts in his last eight trips to the store: $32.92, 14.14, 30.80, 28.34, 75.58, 36.33, 33.51, and 22.94. Which statement about deleting the amount spent of $75.58 is false? a. The median of this data set may be more representative of the mean. b. The value of the mean would be reduced by $5.89. c. The value of the mean would not change. d. The value of the mean would be greatly reduced.
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In a city with three high schools, all the ninth graders took a Standardized Test, with these results: High School Mean Test Score Number of ninth graders (Frequency) Glenwood 79 279 Central 94 349 Lincoln 70 161 The city's PR manager, who never took statistics, claimed the mean score of all ninth graders in the city was the average of 79, 94, and 70, which is 81. Of course, that is incorrect. What is the mean score for all ninth graders in the city? Round to one decimal place. Mean of all ninth grader's Scores =
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