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Question;1. The stem-and-leaf diagram below displays the number of vacation days taken in 1997 by a sample of40 employees of an electronics company.Stem Leaves0 00123334456889 Stem unit = ten1 0001111344562 00001234683 0014 3Referring to the stem-and-leaf display above, what PERCENTAGE of employees took 30 or more vacationdays?A. 10 B. 0.05 C. 5 D. 4 E. 0.10QUESTIONS #2 THROUGH #4 USE THE DATA GIVEN BELOW.A survey was taken to determine how much time UHD students spend doing homework per week. The random sample of UHD students resulted in the frequency distribution shown below.Time (in hours) Frequency0 and under 2 42 and under 4 64 and under 6 116 and under 8 188 and under 10 2610 and under 12 52. The class width is:A. 10.00 B. 2.00 C. 12.00 D. 1.99 E. 1003. What would be the APPROXIMATE shape of the frequency histogram?A. bell-shaped B. skewed to the right C. skewed to the leftD. bi-modal E. none of these4. What PERCENTAGE of the students studied less than 6 hours?A. 11 B. 70 C. 0.16 D. 30 E. 0.35. For a large survey of students, this graph would most likely describe:A. Students? music preference (rock=0, opera=1)B. Students? handed-ness (right=0, left=1)C. Students? political partiesD. Students? heights in inchesQUESTIONS #6 THROUGH #9 USE THE DATA GIVEN BELOW.A random sample of eleven college students were surveyed and asked what brand of laptop they use and how long the battery lasted after the last full charge:Laptop Brand: Samsung, Dell, HP, Apple, Dell, HP, Dell, Toshiba, Dell, HP, AppleBattery Life (hours): 9, 6, 4, 9, 6, 3, 8, 10, 9, 5, 8Assuming this data reflects all students:6. Assuming this data reflects all students at the college, what brand of laptop would you expect the averageor typical student to own?A. Samsung B. HP C. Toshiba D. Apple E. Dell7. A ____________ could be used to display the frequencies of the battery life data, and a ____________could be used to display the frequencies of the laptop brands.A. pie chart, histogram B. bar chart, stem-and-leaf displayC. histogram, bar chart D. stem-and-leaf display, histogramE. histogram, stem and leaf8. What would be the mean time the batteries lasted since the last full charge (in hours)?A. 9 B. 8 C. 7 D. 10 E. 7.59. The sample standard deviation of these battery lives, rounded to the nearest hundredth, would be:A. 2.32 B. 2.65 C. 7.00 D. 5.40 E. 2.22 QUESTIONS #10 THROUGH #13 REFER TO THE FOLLOWING EXCEL OUTPUT:Over the last 6 months, monthly returns (%) were tracked for 4 stock funds.FUND 1 FUND 2 FUND 3 FUND 4January 15 15 24 23February 12 13 15 24March 11 10 13 24April 18 5 9 24May 16 19 19 22June 11 10 13 25FUND 1 FUND 2 FUND 3 FUND 4Mean 13.83333333 12 15.5 23.66666667Standard Error 1.194896555 1.966384161 2.156385865 0.421637021Median 13.5 11.5 14 24Mode 11 10 13 24Standard Deviation 2.926886856 4.816637832 5.282045058 1.032795559Sample Variance 8.566666667 23.2 27.9 1.066666667Kurtosis -1.810019834 0.14343044 0.254750067 0.5859375Skewness 0.388189509 0.032215919 0.71453263 -0.665669013Range 7 14 15 3Minimum 11 5 9 22Maximum 18 19 24 25Sum 83 72 93 142Count 6 6 6 610. Which of these funds carries the most risk with respect to monthly returns?A. Fund 4 B. Fund 3 C. Fund 2 D. Fund 1 E. none of these11. Assuming an approximately normal distribution of return rates for Fund 4, we wouldexpect APPROXIMATELY 95% of the monthly return rates for Fund 4 to be in which of thefollowing intervals?A. (20.57, 26.77) B. (22, 25) C. (22.63, 24.70) D. (23.25, 24.07) E. (21.6, 25.73)12. The 50th percentile of Fund 3?s monthly returns would be ________?A. 14 B. 15 C. 11.5 D. 13 E. 15.513. Consider the January monthly returns for Fund 1 and Fund 2. Relative to their own fund, which Januaryreturn had the better performance?A. They were the same B. Fund 1 C. Fund 2 D. cannot be determinedQUESTIONS #14 AND #15 USE THE TABLE GIVEN BELOW.The table contains the probability distribution for the number of televisions per household in the United States.X P(X)0 .081 .252 .383 .194 .085 .0214. What is the expected number of televisions per household?A. 2.0 B. 2.1 C. 2.5 D. 3.0 E. 2.815. What is the probability that a household has more than 2 televisions?A. 0.71 B. 0.67 C. 0.38 D. 0.33 E. 0.29QUESTIONS #16 AND #17 PERTAIN TO THE FOLLOWING:A study has shown that 30% of all students taking finance at Bellaire College fail the first exam. A random sample of 12 students taking finance at Bellaire College is selected.16. Find the probability that less than 5 students will fail the first finance exam.A. 0.1179 B. 0.1585 C. 0.2764 D. 0.7236 E. 0.882117. On average, how many students would you expect to fail the first finance exam?A. 8.4 B. 3.6 C. 1.5 D. 1.2 E. 1.9018. Determine P(-1.3 < Z 70,000 Ha:?? 70,000D. H0:?? 70,000 Ha:? 0 D. Ho:?1 = 228 vs. Ha:?1 = 228QUESTIONS #36 THROUGH #40 REFER TO THE FOLLOWING:A professor wants to analyze the number of hours students spent online during the weekend and their scores on a test the following Monday. Part of the regression output appears below.Coefficients Standard Error t Stat P-valueIntercept 93.97 4.52 2.77 0.050693Hours Online -4.07 0.86 -4.73 0.008048R Square 0.69105342 36. Interpret the SLOPE of the least squares line.A. For every additional hour spent online, the predicted test score increases by 93.97 points, on average.B. For every additional hour spent online, the predicted test score decreases by 4.07 points, on averageC. For every additional hour spent online, the predicted test score increases by 4.52 points, on average.D. For every 1 point increase in test score, the predicted time spent online decreases by 4.07 hrs, on averageE. For every 1 point increase in test score, the predicted time spent online increases by 93. min. on average.37. Determine the value of the test statistic that the manager used to test whether test score is a usefulpredictor of performance.A. -4.73 B. 4.52 C. 93.97 D. 2.77 E. -4.0738. At the.05 level, the professor could conclude that the time spent online by students over the weekend is auseful predictor of test scores the following Monday.A. TRUEB. FALSE39. What is the least squares regression equation?A. B.C. D.40. _________________is the percentage of variation in performance explained by test score.A. 93.974% B. 86% C. 4.07% D. 69.11% E. 0.69%

 

Paper#62253 | Written in 18-Jul-2015

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