STAT 200 FINAL EXAM 2020 | STAT200 FINAL EXAM_Graded A

STAT 200 FINAL EXAM 2020 Answer all 25 questions. Make sure your answers are as complete as possible. Show all of your work and reasoning. In particular, when there are calculations involved, you must show how you come up with your answers with critical work and/or necessary tables. Answers that come straight from programs or software packages will not be accepted. If you need to use software (for example, Excel) and /or online or hand-held calculators to aid in your calculation, please cite the sources and explain how you get the results. 1. True or False. Justify for full credit. (a) If the variance of a data set is zero, then all the observations in this data set are zero. No, all the observations will be same but it is not necessary that all the observations will be zero, because if all the observations are same even then the variance will be zero. (b) If P(A) = 0.4 , P(B) = 0.5, and A and B are disjoint, then P(A AND B) = 0.9. No, because if A and B are disjoint then P(A AND B) = 0. (c) Assume X follows a continuous distribution which is symmetric about 0. If , then . Yes, P(X<-3) will be less than 0.3 because the value will be towards negative side. (d) A 95% confidence interval is wider than a 90% confidence interval of the same parameter. Yes, a 95% confidence interval is wider than a 90% confidence interval of the same parameter because increasing the confidence limit we increases the acceptance region and reduces the rejection region. (e) In a right-tailed test, the value of the test statistic is 1.5. If we know the test statistic follows a Student’s t-distribution with P(T < 1.5) = 0.96, then we fail to reject the null hypothesis at 0.05 level of significance . No, we will not be able to reject the null hypothesis as we reject the null hypothesis is test statistic is greater than the tabulated value of t. Refer to the following frequency distribution for Questions 2, 3, 4, and 5. Show all work. Just the answer, without supporting work, will receive no credit. The frequency distribution below shows the distribution for checkout time (in minutes) in UMUC MiniMart between 3:00 and 4:00 PM on a Friday afternoon. Checkout Time (in minutes) Frequency Relative Frequency 1.0 - 1.9 3 2.0 - 2.9 12 3.0 - 3.9 0.20 4.0 - 4.9 3 5.0 -5.9 Total 25 2. Complete the frequency table with frequency and relative frequency. Express the relative frequency to two decimal places. Checkout Time (in minutes) Frequency Relative Frequency 1.0 - 1.9 3 0.12 2.0 - 2.9 12 0.48 3.0 - 3.9 5 0.20 4.0 - 4.9 3 0.12 5.0 -5.9 2 0.08 Total 25 1 3. What percentage of the checkout times was at least 3 minutes? Percentage of the checkout times was at least 3 minutes = (5 + 3 +2)/25*100 = 40% 4. In what class interval must the median lie? Explain your answer. We will make the intervals continuous by subtracting 0.05 from lower limit and adding 0.05 in upper limit therefore the corresponding intervals will be 0.95-1.95, 1.95 - 2.95, 2.95 - 3.95, 3.95 - 4.95, 4.95 -5.95. N=25 therefore, N/2=12.5, the immediate less observation lies in 2.0-2.9 i.e. 1.95-2.95. 5. Does this distribution have positive skew or negative skew? Why? The distribution is positively skewed as the frequency in the second interval is 12 and all the proceeding frequencies are very less. This can also be seen in the following graph Frequency 14 12 10 8 6 4 2 0 1.0 - 1.9 2.0 - 2.9 3.0 - 3.9 4.0 - 4.9 5.0 -5.9 Checkout Time (in minutes) Frequency Refer to the following information for Questions 6 and 7. Show all work. Just the answer, without supporting work, will receive no credit. Consider selecting one card at a time from a 52-card deck. (Note: There are 4 aces in a deck of cards) 6. If the card selection is without replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form) (5 pts) The probability that the first card is an ace and the second card is also an ace: P(First card is an Ace and Second card is also an Ace)=(4/52)*(3/51) =0. 7. If the card selection is with replacement, what is the probability that the first card is an ace and the second card is also an ace? (Express the answer in simplest fraction form) The probability that the first card is an ace and the second card is also an ace: P(First card is an Ace and Second card is also an Ace)=(4/52)*(4/52) =0. Refer to the following situation for Questions 8, 9, and 10. The five-number summary below shows the grade distribution of two STAT 200 quizzes for a sample of 500 students.