Final Exam Practice Set, STT 315, Section 106 Options in BOLD are correct choices.: Question 1. Refer following sentences: I. If you flip a FAIR coin many, many times; the proportion of heads will be approximately 0.50 and this proportion will tend to closer to 0.50 as number of tosses increases. II. P( Type I error)= P( Reject Null Hypothesis H 0 Given that H 0 is False). III. R 2 is amount of variability as explained in response variable as explained by independent variable. IV. For a Asymmetric distribution, Mean is always greater than Median. V. Let p=proportion of Toyota or Honda vehicles in East Lansing. Suppose 99% Confidence Interval = (0.48, 0.52). Margin of Error in this case =0.02. Circle the correct Answer. (15 Points) (i) All the above statements are TRUE. (ii) Only statements I, III and V are TRUE. Statements II and IV are TRUE. (iii) Only statements II and III are FALSE. Statements I,IV and V are FALSE. (iv) Only statements I, II and V are TRUE. Statements II and IV are FALSE. (v) None of above options (i), (ii), (iii), (iv) is TRUE. Question 2. A Reporter wants to estimate the proportion of public who support the war in Iraq at his time. He asks 100 people he encounters on the street, this question: Now that almost 4,000 Americans soldiers have died in the war, was the President wise to order an attack in Iraq? Which one is correct? (10 Points) (i) The reporter has used a simple random sample. 1
(ii) The sample is probably too large, since there are only two possible answers. (iii) To avoid any bias, the reporter should balance his survey by also asking a question about the success of the Iraqi election. (iv) He should wear a blindfold while asking a question. (v) The question may bias the answer. Question 3. A Market research company employs a large number of typists to enter the data into a computer database. The time taken for new typist to learn the computer system is known to have a normal distribution with a mean of 130 minutes and standard deviation of 20 minutes. A candidate is automatically hired if he or she learns the computer system in less than 100 minutes. A cut-off time is set to slowest 40% of the learning distribution. Anyone slower than this cut-off time is not hired. Based on this information, answer questions A - C. (A): The proportion of new typist that take under two hours to learn the computer system is: (10 Points) i) 0.159 (ii) 0.309 iii) 0.067 (iv) 0.023 (v) None of these. (B): The proportion candidates that will be automatically hired is: (10 Points) i) 0.023 (ii) 0.159 iii) 0.309 (iv) 0.067 (v) None of these. (C): The cut-off time used by the Market research company is: (10 Points) (i) Two hours and Forty Minutes (ii) Two hours and thirty minutes (iii) Two hours and Fifteen minutes (iv) Two hours and Eight Minutes (v) None of these. Question 4. (A): Which of the following can not be used as measure of spread:(10 Points) (i) Median (ii) Range (iii) Inter Quartile Range (iv) Standard Deviation. (v) None of the above can be used as measure of spread. 2
(B): Two events A and B are such that P (B) = 0.4, P (A c ) = 0.6, P (A c B) = 0.8, then P (A c and B) is: (10 Points) (i) 0.48 (ii) 0.36 (iii) 0.32 (iv) 0.24 (v) None of these. Question 5. Refer the below sentences: I. For a given sample size, a higher confidence means higher margin of Error. II. For a fixed confidence interval level ( i.e. (1-α) is fixed), larger sample means, lower margin of Error. III. For a given confidence level, a sample 16 times as large will make a margin of error one third as small. IV. Suppose n=625; p=0.5, Width of 95% confidence interval level is 0.0788. Circle the correct Answer. (15 Points) (i) Only statements I, II and III are TRUE. Statements IV is TRUE. (ii) All the above statements are TRUE. (iii)only statements I, II and IV are TRUE. Statements II is FALSE. (iv) Only statements III and IV are FALSE. Statements I and II are TRUE. (v) None of above options (i), (ii), (iii), (iv) is TRUE. Question 6. At a large college, all engineering freshmen must take ONE foreign language class, chosen from the languages Spanish, French, Swahili, Chinese and Arabic. The probability distribution for language studied by a randomly selected freshmen is summarized in the following table: Based on this information, answer the questions A-D. Language Spanish French Swahili Chinese Arabic Probability 0.48? 0.09 0.19 0.12 3
(A): The probability that a randomly selected freshmen is studying french is: (5 Points) (i) 0.08 (ii) 0.12 (iii) 0.20 (iv) 0.24 (v) None of these. (B): The probability that a randomly selected freshmen is studying Either Chinese or Arabic is: (i) 1.00 (ii) 0.69 (iii) 0.31 (iv) 0.25 (v) None of these. (C): For a group of three randomly selected freshmen, the probability that All of them studying Spanish is: (i) 1.44 (ii) 0.89 (iii) 0.14 (iv) 0.11 (v) None of these. (D): The one most appropriate display for the above data will be: (i) Histogram (ii) Pie diagram (iii) Bar-diagram (iv) Frequency Distribution (v) None of these. Question 8. (A): Suppose first a word is chosen at random from the sentence We have Final Exam today and then a letter is selected at random from the chosen word. The probability that the letter e will be selected is: i) 0.19 (ii) 0.20 (iii) 0.23 (iv) 0.60 (v) None of these. (B):. Traffic checks on a certain sections of highway suggests that 70% of the drivers are speeding there. Since 0.7*0.7=0.49. Therefore: Reasoning : The Multiplication Rule of Probability might suggest that there is 49% chance 4
that two vehicle on road are speeding. Assumption (I) The two vehicle running on road are disjoint events. Assumption (II) The two vehicles running on the road are Mutually Exclusive Events. Choose the correct answer: (i) Only Assumption (I) is sufficient to make the reasoning TRUE. (ii) Only Assumption (II) is sufficient to make the reasoning TRUE. (iii) Either of Assumption (I) or Assumption (II) is sufficient to make the reasoning TRUE. (iv) None of the above assumptions are enough to make the reasoning true. (v) There is not enough information in the above question. Question 9. One tailed or Two tailed hypothesis? I. A Business students conducts a test to see whether students prefer Diet Coke or Pepsi. II. PepsiCo recently reformulated Diet Pepsi in an attempt to appeal to teenagers. They run a test to see if the new formula appeals to teenagers more than the standard formula. III. A budget override in a small town requires a two thirds majority to pass. newspaper conducts a poll to see if the there is an evidence it will pass. A local IV. A pharmaceutical company is interested if the new drug has improved performance over the old drugs. Circle the correct Answer. (10 Points) (i) Only Hypothesis I is two tailed. (ii) All the above Hypothesis are Two tailed. (iii) Only Hypothesis III and IV are one tailed. Hypothesis I and II are two tailed. (iv) Only Hypothesis I and II are two sided. (v) None of above options (i), (ii), (iii), (iv) is TRUE. 5
Question 10.. A movie theater offers discounted Advance Purchase tickets to customer who buy tickets more than a week days of show and charges regular fare for tickets purchased on or after the day of show. The movie theater manager notices that 60% of it s customer take advantage of the advance purchase of movie tickets. The no-show rate among the people who paid regular ticket prices is 30%, but only 5% of customers with advance purchase tickets are no-shows. Answer the questions below: (A): The percent of no-show ticket holders is: v (i) 15% (ii) 12% (iii) 1.5% (iv) 18% (v) 3% (B): Probability that a customer who did not show had an advance purchase ticket: (5 Points) (i) 15% (ii) 20% (iii) 8.33% (iv) 50% (v) 16.6% Question 11. A software company bids on two contracts. It anticipates a profit of $40,000 if it gets the larger profit and a profit of $10,000 on the smaller contract. The company estimates that there is 30% chance it will get larger contract and 60% chance of getting a smaller contract. Assuming that contracts will be awarded independently, the expected profit and variance are: (10 Points) Solution. Here the prob distribution is given by: Let X be profit. X 0 10000 40000 50000 P(X) 0.28 0.42 0.12 0.18 i) $18,000 and $14,696.93 respectively. ii) $14696.93 and $18,0009 respectively. iii) $18,000 and $215,999,851.4 respectively. iv) $18,000 and $215,999,751.4 respectively. v) None of these. Question 12. Does the fidgeting keeps you slim? Some people don t gain weight even when they overeat. Perhaps fidgeting and other non-exercise activity (NEA) explains why-the 6
body might spontaneously increase non-exercise activity when fed more. Researchers deliberately overfed 16 health young adults for 8 weeks. They measured fat gain in kilograms (the y-variable) and, as an explanatory variable, increase in energy use in the calories from the activity other than deliberate exercise-fidgeting, daily living, and the like ( x variable). Here is the data: The average NEA increase of these 16 adults is 324.75 and the standard deviation is 227.567, NEA increase (x) 94 57 29 135 143 151 245 355 Fat Gain (y) 4.2 3.0 3.7 2.7 3.2 3.6 2.4 1.3 NEA increase (x) 392 473 486 535 571 580 620 690 Fat Gain (y) 3.8 1.7 1.6 2.2 1.0 0.4 2.3 1.1 while the average fat gain is 2.49 with standard deviation 1.15. The correlation between NEA increase and the FAT gain is -0.90. Based on this information, answer the questions below: (A): We want to use a linear model to predict the fat gain by NEA increase. the value of slope b 1 of our linear model? What is (i) -0.004548 (ii) -0.779 (iii) -0.00344 (iv) 0.00735 (B): We want to use a linear model to predict the fat gain by NEA increase. the value of intercept b 0 of our linear model? What is (i) 2.679 (ii) 3.967 (iii) 255.224 (iv) 326.609 (C): Based on the previous linear model, what is the best guess of the amount of the fat gain if the NEA increase is 200? (i) 5.0888 (ii) 3.0574 (iii) 8.659 (iv) 701.021 (D): What percent of variation in fat is explained by variation in NEA increase? (i) 90.2% (ii) -90.0% (iii) 60.7% (iv) 81.0% 7
8